## Basic Trigonometric Identities

In this series, the topic **Introduction to trigonometry** is covered.

This series complements Chapter 8 of NCERT Class 10 Maths and consists of 4 parts:

- Trigonometric ratios,
- Trigonometric ratios for selected angles,
- Trigonometric ratio relations for complementary angles, and,
- Basic trigonometric identities.

This fourth and last part complements **Exercise 4 of Chapter 8 of NCERT Class 10 Maths**, in short, Ex 8.4 class 10.

The session contains, *Concept explanation and problem solutions on Trigonometric Identities.*

This part has **three components:**

**First:**concepts explained in an easy to understand way,**Second:**example problems used for establishing the concepts clearly and,**Third:**the solutions to the problems in the 4th exercise of NCERT Chapter 8.

**To Skip the Concept Explanations** and * Go straight to the Solutions*,

**click here.**### What are basic trigonometric identities

First let us recall the concept of a mathematical identity,

If an equation is true for all values of the variables involved, the equation is called an identity.

In a trigonometric identity,

- Each side of the identity equation will be an expression in trigonometric ratios, and,
- For all values of the variables—the angles and the trigonometric ratio values—the two expressions in LHS and RHS will be equal.

In earlier parts of this chapter, we have mentioned about the first and most basic trigonometric identity—$\sin^2 A+\cos^2 A=1$.

Now we'll prove the truth of the identity and derive more basic trigonometric identities.

#### Proof of $\sin^2 A+\cos^2 A=1$

The proof of this equation is very simple and you can do it yourself.

We'll use the right-angled $\triangle ABC$ in the following diagram for explanation.

First we'll use the geometric resource of Pythagoras theorem to express the sum of squares of **height** BC and **base** AB as equal to the square of **hypotenuse** CA,

$BC^2+AB^2=CA^2$. $\qquad \qquad \qquad..........................(1)$

Just divide both sides of the equation by $CA^2$,

$\left(\displaystyle\frac{BC}{CA}\right)^2+\left(\displaystyle\frac{AB}{CA}\right)^2=1$.$\qquad \qquad..........................(2)$

In equation (2), apply trigonometric ratio concepts to replace,

$\displaystyle\frac{BC}{CA}=\sin A$, and,

$\displaystyle\frac{AB}{CA}=\cos A$.

Result is,

$\sin^2 A+\cos^2 A=1$. $\qquad \qquad \qquad..........................(3)$

This is the first trigonometric identity and it is true for all values in the range, $0^0 \leq \angle A \leq 90^0$. Check it out yourself.

#### Use of the identity $\sin^2 A+\cos^2 A=1$ in various ways

The basic form of the first identity is,

$\sin^2 A+\cos^2 A=1$.

Time again though you may need to use the following variations derived from this identity,

$\sin^2A+\cos^2=1$,

Or, $\sin^2 A=1-\cos^2 A$, $\qquad \qquad..........................(3.1)$

Or, $\cos^2 A=1-\sin^2 A$. $\qquad \qquad..........................(3.2)$

Also an important pair of relation you may often encounter that use this first identity,

$\sin^2 A - \cos^2 A=1-2\cos^2 A$ $\qquad.......................(3.3)$

$\cos^2 A-\sin^2 A=1-2\sin^2 A$ $\qquad.......................(3.4)$

All these four variations also are trigonometric identities.

#### Proof of the second trigonometric identity $1+\tan^2 A=\text{sec}^2 A$

You will get the second basic trigonometric identity by dividing the LHS and RHS of equation (1) by base $AB$ and not by the hypotenuse $CA$.

The transformed equation becomes,

$BC^2+AB^2=CA^2$,

Or, $\left(\displaystyle\frac{BC}{AB}\right)^2+1=\left(\displaystyle\frac{CA}{AB}\right)^2$,

Or, $1 +\tan^2 A=\text{sec}^2 A$.$\qquad \qquad..........................(4)$

You already know that, $\tan A=\displaystyle\frac{BC}{AB}$ and $\text{sec }A=\displaystyle\frac{CA}{AB}$.

#### Valid range of $\angle A$ values for which $1+\tan^2 A=\text{sec}^2 A$ is true

As at $\angle A=90^0$, $\cos A=0$ in the denominator, $\tan A$ at $\angle A=90^0$ is undefined, and so this second basic identity is undefined at $\angle A=90^0$.

At $\angle A=0^0$, $\cos A=\text{sec }A=1$, and $\tan A=0$. So this second identity is valid for $\angle A=0^0$.

Thus the valid range of values for this second identity $1+\tan^2 A=\text{sec}^2 A$ is $0^0 \leq \angle A \lt 90^0$.

#### Use of $1+\tan^2 A=\text{sec}^2 A$ in various innovative ways

The original form of this second trigonometric identity can be used in a number of innovative ways.

$1 +\tan^2 A=\text{sec}^2 A$,

Or, $\tan^2 A=\text{sec}^2 A-1$, $\qquad \qquad \qquad..........................(4.1)$

Or, $\text{sec}^2 A-\tan^2 A=1$. $\qquad \qquad \qquad..........................(4.2)$

This second variation has its LHS in the form of $(a^2-b^2)$ and so it can be split up into two factors,

$\text{sec}^2 A-\tan^2 A=1$,

Or, $(\text{sec} A-\tan A)(\text{sec} A+\tan A)=1$ $ \qquad.....................(4.3)$

Or, $\text{sec} A-\tan A=\displaystyle\frac{1}{\text{sec} A+\tan A}$, $\qquad \qquad....................(4.4)$

Or, $\text{sec} A+\tan A=\displaystyle\frac{1}{\text{sec} A-\tan A}$. $\qquad \qquad....................(4.5)$

The last two variations essentially are mutually inverse relations between the two factors, and hold great power to simplify complex trigonometric expressions quickly and easily. These are **innovative special forms of the second basic trigonometric identity.**

These variations of the second basic trigonometric identity are what we call **advanced trigonometric identities.**

#### Proof of the third trigonometric identity of $1+\text{cot}^2 A=\text{cosec}^2 A$

The third basic basic trigonometric identity is obtained by dividing the LHS and RHS of equation (1) by height $BC$ and not by the hypotenuse $CA$, or base $AB$.

The transformed equation becomes,

$BC^2+AB^2=CA^2$,

$1+\left(\displaystyle\frac{AB}{BC}\right)^2=\left(\displaystyle\frac{CA}{BC}\right)^2$,

$1+\text{cot}^2 A=\text{cosec}^2 A$.$\qquad \qquad..........................(5)$

This is the third basic trigonometric identity.

#### Valid range of $\angle A$ values for which $1+\text{cot}^2 A=\text{cosec}^2 A$ is true

As at $\angle A=0^0$, $\sin A=0$ in the denominator, $\text{cot} A$ at $\angle A=0^0$ is undefined, and so the second identity is undefined at $\angle A=0^0$.

At $\angle A=90^0$, $\sin A=\text{cosec }A=1$, and $\text{cot} A=0$. So this third basic identity is valid for $\angle A=90^0$.

Thus the valid range of values for this third basic identity $1+\text{cot}^2 A=\text{cosec}^2 A$ is $0^0 \lt \angle A \leq 90^0$.

#### Use of $1+\text{cot}^2 A=\text{cosec}^2 A$ in various innovative ways

The original form of this third basic trigonometric identity can be used in a number of innovative ways.

$1 +\text{cot}^2 A=\text{cosec}^2 A$,

Or, $\text{cot}^2 A=\text{cosec}^2 A-1$, $\qquad \qquad \qquad.......................(5.1)$

Or, $\text{cosec}^2 A-\text{cot}^2 A=1$. $\qquad \qquad \qquad.......................(5.2)$

This second variation has its LHS in the form of $(a^2-b^2)$ and so it can be split up into two factors,

$\text{cosec}^2 A-\text{cot}^2 A=1$,

Or, $(\text{cosec} A-\text{cot }A)(\text{cosec} A+\text{cot }A)=1$ $ \qquad...............(5.3)$

Or, $\text{cosec} A-\text{cot }A=\displaystyle\frac{1}{\text{cosec} A+\text{cot }A}$, $\qquad \qquad..............(5.4)$

Or, $\text{cosec} A+\text{cot }A=\displaystyle\frac{1}{\text{cosec} A-\text{cot }A}$. $\qquad \qquad..............(5.5)$

The last two variations essentially are mutually inverse relations between the two factors, and hold great power to simplify complex trigonometric expressions quickly and easily. These are **innovative special forms of the third basic trigonometric identity.**

These variations of the third basic trigonometric identity are what we call **advanced trigonometric identities.**

**Note:** In the earlier three parts, you have learned,

- how to find values of all trigonometric ratios when value of a single trigonometric ratio is given by ratios of two sides of a right-angled triangle, and
- how to simplify trigonometric expressions using trigonometric ratio values for selected angles in a limited way.

Both these you could do only in very limited ways.

The **real power of trigonometry problem solving comes from using the trigonometric identities.**

Just as mentioned above, basic trigonometric identities are used in two ways for solving problems,

- finding values of all other trigonometric ratio functions when numeric value of a single trigonometric ratio function is given, and more importantly,
- simplifying all types of simple and complex trigonometric expressions, usually identities themselves, using the basic trigonometric identities.

We'll now show these two types of uses by solving a few problem examples.

### Examples on solving problems using basic trigonometric identities

#### Problem example 1.

Express $\sin A$, $\text{cot }A$, and $\text{cosec }A$ in terms of $\cos A$, and find their values, when $\cos A=\displaystyle\frac{1}{\sqrt{3}}$, and $0^0 \lt \angle A \leq 90^0$.

#### Solution example 1.

First basic trigonometric identity is,

$\sin^2 A+\cos^2 A=1$,

Or, $\sin A=\sqrt{1-\cos^2 A}=\sqrt{\displaystyle\frac{2}{3}}$, $\sin A$ can't be negative as $\angle A$ is an acute angle.

So,

$\text{cot }A=\displaystyle\frac{\cos A}{\sin A}=\displaystyle\frac{\cos A}{\sqrt{1-\cos^2 A}}$

$=\displaystyle\frac{\displaystyle\frac{1}{\sqrt{3}}}{\displaystyle\frac{\sqrt{2}}{\sqrt{3}}}=\frac{1}{\sqrt{2}}$, and

$\text{cosec }A=\displaystyle\frac{1}{\sin A}=\displaystyle\frac{1}{\sqrt{1-\cos^2 A}}=\sqrt{\displaystyle\frac{3}{2}}$.

#### Problem example 2.

Prove that $\text{cosec } A(1-\cos A)(\text{cosec }A+\text{cot }A)=1$.

#### Solution example 2.

In the LHS of the given equation, first multiply $\text{cosec }A$ through the first brackets of $(1-\cos A)$. The transformed LHS is,

$\text{cosec } A(1-\cos A)(\text{cosec }A+\text{cot }A)$

$=\left(\text{cosec }A-\displaystyle\frac{\cos A}{\sin A}\right)(\text{cosec }A+\text{cot }A)$

$=(\text{cosec }A-\text{cot }A)(\text{cosec }A+\text{cot }A)$

$=\text{cosec}^2 A-\text{cot}^2 A$, using algebraic identity, $(a-b)(a+b)=a^2-b^2$

$=1+\text{cot}^2 A-\text{cot}^2 A$, using basic trigonometric identity $1+\text{cot}^2 A=\text{cosec}^2 A$

$=1$. Proved.

#### Problem example 3.

Prove that $\displaystyle\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\text{sec }A-\tan A}$.

#### Solution example 3.

Observing $\text{sec }A$ and $\tan A$ on the RHS, decide to simply divide both the numerator and denominator of LHS by $\cos A$ so that the LHS also is converted in terms of $\text{sec }A$ and $\tan A$.

LHS of the given equation is thus transformed to,

$LHS=\displaystyle\frac{\sin A-\cos A+1}{\sin A+\cos A-1}$

$=\displaystyle\frac{\tan A-1+\text{sec }A}{\tan A+1-\text{sec }A}$

$=\displaystyle\frac{(\text{sec }A+\tan A)-1}{\tan A+1-\text{sec }A}$

**To create $(\text{sec }A-\tan A)$ in the denominator as in the RHS**, now multiply both the numerator and denominator of the transformed LHS with $(\text{sec }A-\tan A)$.

Result is,

$LHS=\displaystyle\frac{\text{sec }A-\tan A}{\text{sec }A-\tan A}\times{\displaystyle\frac{(\text{sec }A+\tan A)-1}{\tan A+1-\text{sec }A}}$

$=\displaystyle\frac{1}{\text{sec }A-\tan A}\times{\displaystyle\frac{(\text{sec }^2 A-\tan^2 A)-\text{sec }A+\tan A}{\tan A+1-\text{sec }A}}$

$=\displaystyle\frac{1}{\text{sec }A-\tan A}\times{\displaystyle\frac{1-\text{sec }A+\tan A}{\tan A+1-\text{sec }A}}$, as $\text{sec}^2 A-\tan^2 A=1$

$=\displaystyle\frac{1}{\text{sec }A-\tan A}\times{\displaystyle\frac{\tan A+1-\text{sec }A}{\tan A+1-\text{sec }A}}$

$=\displaystyle\frac{1}{\text{sec }A-\tan A}$.

**Proved.**

By **mathematical reasoning** it could be assumed with confidence that after multiplying and dividing the transformed LHS by $(\text{sec }A-\tan A)$, if this multiplying factor is only multiplied with the numerator and kept independent in the denominator, the rest of the resultant numerator and denominator **must become same thus cancelling each other.**

We'll solve now the problems in the exercise.

This is the third and last component of the session.

### Problems to solve: Exercise 4 Chapter 8 NCERT solutions for Class 10 maths: Basic trigonometric identities

#### Problem 1.

Express $\sin A$, $\text{sec }A$, and $\tan A$ in terms of $\text{cot }A$.

#### Solution to Problem 1.

You'll get $\text{cosec }A$ and so $\sin A$ from $\text{cot }A$ by using the identity,

$1+\text{cot}^2 A=\text{cosec}^2 A$

Or, $\text{cosec }A=\sqrt{1+\text{cot}^2 A}$

Or, $\sin A=\displaystyle\frac{1}{\sqrt{1 +\text{cot}^2 A}}$.

Now you'll get $\tan A$ by just inverting $\text{cot }A$,

$\tan A=\displaystyle\frac{1}{\text{cot }A}$.

And finally $\text{sec }A$ from $\tan A$ by using the basic identity,

$\text{sec}^2 A=1+\tan^2 A=1+\displaystyle\frac{1}{\text{cot}^2 A}$

$=\displaystyle\frac{1+\text{cot}^2 A}{\text{cot}^2 A}$,

Or, $\text{sec }A=\displaystyle\frac{\sqrt{1+\text{cot}^2 A}}{\text{cot }A}$.

**Answer:** $\sin A=\displaystyle\frac{1}{\sqrt{1 +\text{cot}^2 A}}$, $\text{sec }A=\displaystyle\frac{\sqrt{1+\text{cot}^2 A}}{\text{cot }A}$, and $\tan A=\displaystyle\frac{1}{\text{cot }A}$.

#### Problem 2.

Write all the other trigonometric ratios of $\angle A$ in terms of $\text{sec }A$.

#### Solution to Problem 2.

The first and easiest one to get is,

$\cos A=\displaystyle\frac{1}{\text{sec }A}$.

Once you get $\cos A$, next to get is $\sin A$ from the trigonometric identity,

$\sin^2 A+\cos^2 A=1$,

Or, $\sin A=\sqrt{1-\cos^2 A}=\sqrt{1-\displaystyle\frac{1}{\text{sec}^2A}}$

Or, $\sin A=\displaystyle\frac{\sqrt{\text{sec}^2 A-1}}{\text{sec }A}$.

You'll get $\tan A$ easily from $\text{sec }A$ by using the identity,

$1+\tan^2 A=\text{sec}^2 A$,

Or, $\tan A=\sqrt{\text{sec}^2 A-1}$.

Then $\text{cot }A$ as inverse of $\tan A$,

$\text{cot }A=\displaystyle\frac{1}{\sqrt{\text{sec}^2A-1}}$.

Finally $\text{cosec }A$ as inverse of $\sin A$,

$\text{cosec }A=\displaystyle\frac{\text{sec }A}{\sqrt{\text{sec}^2 A-1}}$.

**Answer:** $\sin A=\displaystyle\frac{\sqrt{\text{sec}^2 A-1}}{\text{sec }A}$, $\cos A=\displaystyle\frac{1}{\text{sec }A}$, $\tan A=\sqrt{\text{sec}^2 A-1}$,

$\text{cosec }A=\displaystyle\frac{\text{sec }A}{\sqrt{\text{sec}^2 A-1}}$, and $\text{cot }A=\displaystyle\frac{1}{\sqrt{\text{sec}^2A-1}}$.

#### Problem 3.i.

Evaluate $\displaystyle\frac{\sin^2 63^0+\sin^2 27^0}{\cos^2 17^0+\cos^2 73^0}$.

#### Solution to Problem 3.i.

By complementary angle trigonometric ratio relations,

$\sin 27^0=\cos(90^0-27^0)=\cos 63^0$, and

$\cos 17^0=\sin (90^0-17^0)=\sin 73^0$.

Substituting these two results into the given expression you get,

$\displaystyle\frac{\sin^2 63^0+\sin^2 27^0}{\cos^2 17^0+\cos^2 73^0}$

$=\displaystyle\frac{\sin^2 63^0+\cos^2 63^0}{\sin^2 73^0+\cos^2 73^0}$.

As $\sin^2 \theta +\cos^2 \theta=1$,

$\sin^2 63^0+\cos^2 63^0=1$, and,

$\sin^2 73^0+\cos^2 73^0=1$.

So the given expressions evaluates to simply 1.

**Answer:** 1.

#### Problem 3.ii.

Evaluate $\sin 25^0\cos 65^0+\cos 25^0 \sin 65^0$.

#### Solution to Problem 3.ii.

By complementary angle trigonometric ratio relationships,

$\cos 65^0=\cos (90^0-65^0)=\sin 25^0$, and,

$\sin 65^0=\cos (90^0-65^0)=\cos 25^0$.

Substituting these two results in the given expression you get,

$\sin 25^0\cos 65^0+\cos 25^0 \sin 65^0=\sin^2 25^0+\cos^2 25^0=1$,

by the trigonometric identity, $\sin^2 \theta+\cos^2 \theta=1$.

**Answer:** 1.

#### Problem 4.i.

Choose the correct option. Justify your choice.

$9\text{sec}^2 A-9\tan^2 A=$

$\text{a. }1 \qquad \qquad \text{b. }9 \qquad \qquad \text{c. }8 \qquad \qquad \text{d. }0$.

#### Solution to Problem 4.i.

The trigonometric identity between $\text{sec }\theta$ and $\tan \theta$ is,

$1+\tan^2 \theta=\text{sec}^2 \theta$,

Or, $\text{sec}^2 \theta-\tan^2\theta =1$.

Using this identity on the given expression,

$9\text{sec}^2 A-9\tan^2 A=9(\text{sec}^2 A-\tan^2 A)=9$.

**Answer:** Option b. 9.

#### Problem 4.ii.

Choose the correct option. Justify your choice.

$(1+\tan \theta+\text{sec }\theta)(1+\text{cot }\theta-\text{cosec }\theta)=$

$\text{a. }0 \qquad \qquad \text{b. }1 \qquad \qquad \text{c. }2 \qquad \qquad \text{d. }-1$.

#### Solution to Problem 4.ii.

Discover the pattern that if you convert $\tan \theta$ in the first factor expression to $\displaystyle\frac{1}{\text{cot }\theta}$, the factor becomes, $(1+\text{cot }\theta+\text{cosec }\theta)$,

$(1+\tan \theta+\text{sec }\theta)(1+\text{cot }\theta-\text{cosec }\theta)$

$=(1+\displaystyle\frac{1}{\text{cot }\theta}+\text{sec }\theta)(1+\text{cot }\theta-\text{cosec }\theta)$

$=\displaystyle\frac{(1+\text{cot }\theta +\text{cosec } \theta)(1+\text{cot }\theta-\text{cosec }\theta)}{\text{cot } \theta}$

$=\displaystyle\frac{(1+\text{cot }\theta)^2-(\text{cosec }\theta)^2}{\text{cot }\theta}$

$=\displaystyle\frac{1+2\text{cot }\theta+\text{cot}^2 \theta-\text{cosec}^2 \theta}{\text{cot}\theta}$

$=\displaystyle\frac{\text{cosec}^2 \theta+2\text{cot }\theta-\text{cosec}^2 \theta}{\text{cot }\theta}$, as $1+\text{cot}^2 \theta=\text{cosec}^2 \theta$

$=\displaystyle\frac{2\text{cot }\theta}{\text{cot }\theta}$

$=2$.

**Answer:** Option c. 2.

#### Problem 4.iii.

Choose the correct option. Justify your choice.

If $(\text{sec }A+\tan A)(1-\sin A)=$

$\text{a. }\text{sec }A \qquad \qquad \text{b. }\sin A \qquad \qquad \text{c. }\text{cosec }A \qquad \qquad \text{d. }\cos A$.

#### Solution to Problem 4.iii.

Converting the first factor by substituting, $\text{sec }A=\displaystyle\frac{1}{\cos A}$ and $\tan A=\displaystyle\frac{\sin A}{\cos A}$, the given expression is transformed to,

$(\text{sec }A+\tan A)(1-\sin A)$

$=\displaystyle\frac{(1+\sin A)(1- \sin A)}{\cos A}$

$=\displaystyle\frac{1-\sin^2 A}{\cos A}$

$=\displaystyle\frac{\cos^2 A}{\cos A}$, as $\sin^2 A+\cos^2 A=1$, $1-\sin^2 A=\cos^2 A$

$=\cos A$.

**Answer:** Option d. $\cos A$.

#### Problem 4.iv.

Choose the correct option. Justify your choice.

$\displaystyle\frac{1+\tan^2 A}{1+\text{cot}^2 A}=$

$\text{a. }\text{sec}^2 A \qquad \qquad \text{b. }-1 \qquad \qquad \text{c. }\text{cot}^2 A \qquad \qquad \text{d. }\tan^2 A$.

#### Solution to Problem 4.iv.

Using identity, $1+\tan^2 A=\text{sec}^2 A$ in the numerator, and, $1+\text{cot}^2 A=\text{cosec}^2 A$ in the denominator,

$\displaystyle\frac{1+\tan^2 A}{1+\text{cot}^2 A}$

$=\displaystyle\frac{\text{sec}^2 A}{\text{cosec}^2 A}$

$=\tan^2 A$.

**Answer:** Option d. $\tan^2 A$.

#### Strategies and techniques for solving trigonometric identity problems

The following 10 problems in question 5 are all identity proving problems. This type of problems form the most important component of trigonometric problem solving in schools, in class 10 as well as in class 11.

In addition to the knowledge of the three basic trigonometric identities that you have learned till now, you need to adopt effective problem solving strategies and techniques to solve these problems easily and elegantly without trying out various ways you can solve randomly.

These problems are invariably of the form,

Prove that $\text{LHS expression}=\text{RHS expression}$.

The powerful and effective strategies and techniques that we would use for solving these problems follow.

**LHS to RHS processing:** For all such problems, our target would be to form the RHS expression starting from the LHS expression—for any mathematical proving problem, this is the ethical or decent way to prove. We won't simplify both the LHS expression and RHS expression to equate them, for any problem.

**Target driven problem solving:** Target being the RHS expression, on analyzing the target expression only we would take the **most suitable** action to transform the LHS expression so that we would reach the target expression as simply as possible. This is what we call—target driven problem solving. This is a natural way to solve any type of problem.

**Denominator equalization:** When the LHS has more than one term of fraction expressions, our first attempt would be to make the denominator expressions equal, so that the fraction terms can be combined into a single fraction term.

**Denominator elimination:** When the RHS is an expression that is not a fraction with numerator and denominator expression, but the LHS is a fraction expression, primary objective would be to eliminate the denominator of the LHS fraction expression.

**Friendly trigonometric ratio pairs:** We have named $\sin \theta$ and $\cos \theta$ as a friendly trigonometric ratio pair, because in solving a problem involving only $\sin \theta$ and $\cos \theta$, we can use the trigonometric identity $\sin ^2 \theta+\cos^2 \theta=1$ very effectively.

In the same way, $\text{sec }\theta$ and $\tan \theta$ form the second friendly trigonometric ratio pair, as in solving problems involving only these two, we would be able to use the identity, $1+\tan^2 \theta=\text{sec}^2 \theta$.

Likewise, $\text{cosec }\theta$ and $\text{cot }\theta$ form the third friendly trigonometric ratio pair for which we would be able to use the identity, $1+\text{cot}^2 \theta=\text{cosec}^2 \theta$.

**Mutually inverse ratios:** A good example of a problem involving mutually inverse trigonometric ratios is a problem consisting of only $\tan \theta$ and $\text{cot }\theta$.

For solving such a problem, it would always be simpler and faster if you express, say $\text{cot } \theta$ as $\displaystyle\frac{1}{\tan \theta}$, to reduce the whole problem expression in terms of only one variable, that is, $\tan \theta$.

Instead, if you convert the $\tan \theta$ and $\text{cot }\theta$ in terms of $\sin \theta$ and $\cos \theta$ it would lead you to a more cumbersome solution invariably.

Let us solve the 10 problems now to understand how we can use these stategices and techniques in the best possible way.

#### Problem 5.i.

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

$(\text{cosec }\theta-\text{cot }\theta)^2=\displaystyle\frac{1-\cos \theta}{1+\cos \theta}$.

#### Solution to Problem 5.i.

**Apply the powerful problem solving technique**—in LHS express one factor of the two (as it is a square, there are two equal factors) and express it as an inverse and simplify.

Let us show in practice.

The LHS of the given expression,

$LHS=(\text{cosec }\theta-\text{cot }\theta)^2$

$=(\text{cosec }\theta-\text{cot }\theta)(\text{cosec }\theta-\text{cot }\theta)$

$=\displaystyle\frac{\text{cosec }\theta-\text{cot }\theta}{\text{cosec }\theta+\text{cot }\theta}$.

Let us explain.

The identity between $\text{cosec }\theta$ and $\text{cot }\theta$ is,

$1+\text{cot}^2 \theta=\text{cosec}^2 \theta$

Or, $\text{cosec}^2 \theta-\text{cot}^2 \theta=1$,

Or, $(\text{cosec } \theta+\text{cot } \theta)(\text{cosec } \theta-\text{cot } \theta)=1$, as $a^2-b^2=(a+b)(a-b)$,

Or, $\text{cosec } \theta-\text{cot } \theta=\displaystyle\frac{1}{\text{cosec } \theta+\text{cot } \theta}$.

The mutually inverse relationship is very effective in solving apparently hard problems.

Comng back to our problem,

$LHS=\displaystyle\frac{\text{cosec }\theta-\text{cot }\theta}{\text{cosec }\theta+\text{cot }\theta}$

$=\displaystyle\frac{1-\cos \theta}{1+\cos \theta}$, cancelling out $\sin \theta$

$=RHS$

**Proved.**

#### Problem 5.ii.

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

$\displaystyle\frac{\cos A}{1+\sin A}+\displaystyle\frac{1+\sin A}{\cos A}=2 \text{sec }A$.

#### Solution to Problem 5.ii.

You can look ahead to mentally cross-multiply the denominators with opposite numerators of the LHS to form the transformed expression as,

$LHS=\displaystyle\frac{\cos A}{1+\sin A}+\displaystyle\frac{1+\sin A}{\cos A}$

$=\displaystyle\frac{\cos^2 A+(1+\sin A)^2}{\cos A(1+\sin A)}$

$=\displaystyle\frac{2+2\sin A}{\cos A(1+\sin A)}$, as $\sin^2 A+\cos^2 A=1$

$=\displaystyle\frac{2(1+\sin A)}{\cos A(1+\sin A)}$

$=2\text{sec }A$

$=RHS$.

**Proved.**

#### Problem 5.iii.

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

$\displaystyle\frac{\tan \theta}{1-\text{cot }\theta}+\displaystyle\frac{\text{cot } \theta}{1-\tan\theta}=1+\text{sec }\theta \text{cosec }\theta$.

#### Solution to Problem 5.iii.

Instead of converting to $\sin \theta$ and $\cos \theta$ replace $\text{cot }\theta$ in the second term by $\displaystyle\frac{1}{\tan \theta}$.

The first term is kept unchanged because we could see that by the substitution, the two denominators can be equalized easily.

This is application of denominator equalization technique.

The result is,

$LHS=\displaystyle\frac{\tan \theta}{1-\text{cot }\theta}+\displaystyle\frac{\text{cot } \theta}{1-\tan\theta}$

$=\displaystyle\frac{\tan \theta}{1-\text{cot }\theta}-\displaystyle\frac{\text{cot}^2 \theta}{1-\text{cot }\theta}$

$=\displaystyle\frac{\tan \theta-\text{cot}^2 \theta}{1-\text{cot }\theta}$.

Now only replace the single $\tan \theta$ by $\displaystyle\frac{1}{\text{cot }\theta}$, to reduce the expression to a single variable one,

$LHS=\displaystyle\frac{1-\text{cot}^3 \theta}{\text{cot }\theta(1-\text{cot }\theta)}$.

Numerator is in the form of $(1-a^3)=(1-a)(1+a+a^2)$ where $a=\text{cot }\theta$.

Expanding the subtractive sum of cubes in the numerator you would get $(1-\text{cot }\theta)$ cancelled out between the numerator and the denominator leaving,

$LHS=\displaystyle\frac{1+\text{cot }\theta+\text{cot}^2 \theta}{\text{cot }\theta}$.

Now apply the identity, $1+\text{cot}^2 \theta=\text{sec}^2 \theta$,

$LHS=\displaystyle\frac{\text{cot }\theta+\text{sec}^2 \theta}{\text{cot }\theta}$

$=1+\text{sec}^2 \theta \text{cot }\theta$

$=1+\text{sec }\theta \text{cosec }\theta$

$=RHS$.

**Proved.**

You may try yourself solving the problem by breaking up $\tan \theta$ and $\text{cot }\theta$ in terms of $\sin \theta$ and $\cos \theta$.

#### Problem 5.iv.

$\displaystyle\frac{1+\text{sec }A}{\text{sec }A}=\displaystyle\frac{\sin^2 A}{1-\cos A}$.

#### Solution to Problem 5.iv.

We would solve the problem quickly by target driven problem solving strategy.

As an answer to the question—how easily can we create the denominator of $(1-\cos A)$ of the target expression, when we look at the LHS, answer becomes clear.

Just multiply both numerator and denominator of LHS by $(\text{sec }A-1)$. Result you would get,

$LHS=\displaystyle\frac{1+\text{sec }A}{\text{sec }A}=\displaystyle\frac{\text{sec}^2 A-1}{\text{sec }A(\text{sec }A-1)}$

$=\displaystyle\frac{\tan^2 A \cos A}{\text{sec }A-1}$, using identity, $\text{sec}^2 A-1=\tan^2 A$ from $1+\tan^2 A=\text{sec}^2 A$

$=\displaystyle\frac{\tan^2 A \cos^2 A}{1-\cos A}$

$=\displaystyle\frac{\sin^2 A}{1-\cos A}$

$=RHS$.

**Proved.**

#### Problem 5.v.

$\displaystyle\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\text{cosec }A+\text{cot }A$.

#### Solution to Problem 5.v.

As the RHS is in terms of the friendly trigonometric ratio pair, $\text{cosec }A$ and $\text{cot }A$, we would first convert the LHS in terms of these two.

As decided strategically, divide the numerator and denominator by $\sin A$ to transform the LHS to,

$LHS=\displaystyle\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$

$=\displaystyle\frac{\text{cot }A-1+\text{cosec }A}{\text{cot }A+1-\text{cosec }A}$

$=\displaystyle\frac{\text{cot }A-1+\text{cosec }A}{1-(\text{cosec }A-\text{cot }A)}$.

**Our objective is clear**—to form $\text{cosec }A+\text{cot }A$ as RHS, if we multiply both the numerator and denominator of the transformed LHS keeping the multiplying factor independent in the numerator, the product of this factor with exising denominator must result in the existing numerator to cancel it out leaving only the multiplying factor in the numerator as the desired target result.

This is a mathematical certainty and we take this action as decided.

Result is,

$LHS=\displaystyle\frac{\text{cot }A-1+\text{cosec }A}{1-(\text{cosec }A-\text{cot }A)}$

$=\displaystyle\frac{(\text{cosec }A+\text{cot }A)}{(\text{cosec }A+\text{cot }A)}\times{\displaystyle\frac{(\text{cot }A-1+\text{cosec }A)}{[1-(\text{cosec }A-\text{cot }A)]}}$

$=\displaystyle\frac{(\text{cosec }A+\text{cot }A)(\text{cot }A-1+\text{cosec }A)}{\text{cosec }A+\text{cot }A-(\text{cosec}^2 A-\text{cot}^2 A)}$

$=\displaystyle\frac{(\text{cosec }A+\text{cot }A)(\text{cot }A-1+\text{cosec }A)}{\text{cosec }A+\text{cot }A-1}$.

As identity $1+\text{cot}^2 A=\text{cosec}^2 A$, it follows, $\text{cosec}^2 A-\text{cot}^2 A=1$.

So,

$LHS=\displaystyle\frac{(\text{cosec }A+\text{cot }A)(\text{cot }A-1+\text{cosec }A)}{\text{cot }A -1+\text{cosec }A}$

$=\text{cosec }A+\text{cot }A$

$=RHS$.

**Proved.**

#### Problem 5.vi.

$\sqrt{\displaystyle\frac{1+\sin A}{1-\sin A}}=\text{sec }A+\tan A$.

#### Solution to Problem 5.vi.

**Primary objective** must be to **get rid of the square root, as the RHS is without any square root**.

Multiplying both the numerator and denominator within the square root by either $(1+\sin A)$ or $(1-\sin A)$ would achieve this result.

**Second objective**, again driven by the nature of the target expression, is to convert the denominator of the LHS to $\cos A$ which is the common denominator of both the RHS terms.

That decides the final action—we have to multiply both numerator and denominator within the square root by $(1+\sin A)$.

Result is,

$LHS=\sqrt{\displaystyle\frac{1+\sin A}{1-\sin A}}$

$=\sqrt{\displaystyle\frac{(1+\sin A)^2}{1-\sin^2 A}}$

$=\displaystyle\frac{1+\sin A}{\cos A}$

by using the identity $1-\sin^2 A=\cos^2 A$ derived from $\sin^2 A+\cos^2 A=1$.

So,

$LHS=\displaystyle\frac{1+\sin A}{\cos A}$

$=\text{sec }A+\tan A$

$=RHS$.

**Proved.**

#### Problem 5.vii.

$\displaystyle\frac{\sin \theta-2\sin^3 \theta}{2\cos^3 \theta -\cos \theta}=\tan \theta$.

#### Solution to Problem 5.vii.

Factoring out $\sin \theta$ from numerator and $\cos \theta$ from denominator, the LHS is transformed to,

$LHS=\displaystyle\frac{\sin \theta-2\sin^3 \theta}{2\cos^3 \theta -\cos \theta}$

$=\displaystyle\frac{\sin \theta(1-2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta-1)}$

$=\tan \theta\left[\displaystyle\frac{1-2\sin^2 \theta}{2-2\sin^2 \theta-1}\right]$, replacing $\cos^2 \theta$ in denominator by $1-\sin^2 \theta$

$=\tan \theta\left[\displaystyle\frac{1-2\sin^2 \theta}{1-2\sin^2 \theta}\right]$

$=\tan \theta$

$=RHS$.

**Proved.**

#### Problem 5.viii.

$(\sin A+\text{cosec }A)^2+(\cos A+\text{sec }A)^2=7+\tan^2 A+\text{cot}^2 A$.

#### Solution to Problem 5.viii.

Solve by straightforward expansion of the squares and applying identities.

LHS of the given expression is,

$(\sin A+\text{cosec }A)^2+(\cos A+\text{sec }A)^2$

$=\sin^2 A +2+\text{cosec}^2 A+\cos^2 A+2+\text{sec}^2 A$

$=4+(\sin^2 A+\cos^2 A)+(1+\text{cot}^2 A)+(1+\tan^2 A)$, using identities, $1+\text{cot}^2 A=\text{cosec}^2A$, and $1+\tan^2 A=\text{sec}^2 A$

$=7+\tan^2 A+\text{cot}^2 A$, using identity $\sin^2 A+\cos^2 A=1$.

$=RHS$.

**Proved.**

#### Problem 5.ix.

$(\text{cosec }A-\sin A)(\text{sec }A-\cos A)=\displaystyle\frac{1}{\tan A+\text{cot }A}$.

#### Solution to Problem 5.ix.

We'll simplify the LHS to its simplest form first that is clearly visible.

In each of the two factors of LHS, the two terms are mutually inverse to each other. Expressing in inverse form,

$LHS=(\text{cosec }A-\sin A)(\text{sec }A-\cos A)$

$=\left(\displaystyle\frac{1}{\sin A}-\sin A\right)\left(\displaystyle\frac{1}{\cos A}-\cos A\right)$

$=\displaystyle\frac{(1-\sin^2 A)(1-\cos^2 A)}{\sin A\cos A}$

$=\displaystyle\frac{\cos^2 A\sin^2 A}{\sin A\cos A}$.

As, $\cos^2 A=1-\sin^2 A$, and $\sin^2 A=1-\cos^2 A$ from $\sin^2 A+\cos^2 A=1$.

So,

$LHS=\displaystyle\frac{\cos^2 A\sin^2 A}{\sin A\cos A}$

$=\sin A\cos A$.

Now multiply and divide this result with $(\tan A+\text{cot A})$. As a result, the denominaor would remain as it is, but the numerator must turn to 1. Let us see the result.

$LHS=\sin A\cos A\times{\displaystyle\frac{\tan A+\text{cot }A}{\tan A+\text{cot }A}}$

$=\displaystyle\frac{\sin^2 A+\cos^2 A}{\tan A+\text{cot }A}$

$=\displaystyle\frac{1}{\tan A+\text{cot }A}$, using identity $\sin^2 A+\cos^2 A=1$.

$=RHS$

**Proved.**

#### Problem 5.x.

$\left(\displaystyle\frac{1+\tan^2 A}{1+\text{cot}^2 A}\right)=\left(\displaystyle\frac{1-\tan A}{1-\text{cot }A}\right)^2=\tan^2 A$.

#### Solution to Problem 5.x.

This in fact is two problems combined into one.

We'll solve the two identities,

i. $\left(\displaystyle\frac{1+\tan^2 A}{1+\text{cot}^2 A}\right)=\tan^2 A$, and,

ii. $\left(\displaystyle\frac{1-\tan A}{1-\text{cot }A}\right)^2=\tan^2 A$.

**Solution to 5.x.i.**

In the LHS, just replace the numerator by using identity $1+\tan^2 A=\text{sec}^2 A$ and the denominator by using the identity, $1+\text{cot}^2 A=\text{cosec}^2 A$,

$LHS=\left(\displaystyle\frac{1+\tan^2 A}{1+\text{cot}^2 A}\right)$

$=\displaystyle\frac{\text{sec}^2 A}{\text{cosec}^2 A}$

$=\displaystyle\frac{\sin^2 A}{\cos^2 A}$

$=\tan^2 A$

$=RHS$.

Proved.

**Solution to 5.x.ii.**

To prove, $\left(\displaystyle\frac{1-\tan A}{1-\text{cot }A}\right)^2=\tan^2 A$.

In the LHS, convert $\tan A=\displaystyle\frac{\sin A}{\cos A}$, and $\text{cot }A=\displaystyle\frac{\cos A}{\sin A}$. The result becomes,

$LHS=\left(\displaystyle\frac{1-\displaystyle\frac{\sin A}{\cos A}}{1-\displaystyle\frac{\cos A}{\sin A}}\right)^2$

$=\left(\displaystyle\frac{\cos A-\sin A}{\sin A-\cos A}\right)^2\displaystyle\frac{\sin^2 A}{\cos^2 A}$

$=\tan^2 A$.

$=RHS$.

Proved.

Combining the two results,

$\left(\displaystyle\frac{1+\tan^2 A}{1+\text{cot}^2 A}\right)=\left(\displaystyle\frac{1-\tan A}{1-\text{cot }A}\right)^2=\tan^2 A$.

**Proved.**

### End note

If you are a beginner and you have to solve the problems on your own, you may not find the problems very easy to solve. But first, you need to carefully go through the solutions understanding the process of solving fully and then you must solve many more identity problems to gain complete mastery.

Overall, we must say, trigonometric identity problem solving is interesting and never too difficult.

### NCERT Solutions for Class 10 Maths

#### Chapter 1: Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Chapter 2: Polynomials

#### Chapter 3: Linear Equations

**NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.**

#### Chapter 4: Quadratic equations

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square**

#### Chapter 6: Triangles

**NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons**

**Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles**

#### Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities*

#### Chapter 8: Introduction to Trigonometry, only solutions to selected problems

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**