## Problem solving approach gives us the solutions in a few steps

In this session, we will solve again two selected NCERT Trigonometry problems of class 10 standard in a few steps.

In the process, we will **highlight the problem solving approach**, in which, by applying **useful pattern identification, time tested trigonometric and algebraic expression simplifying strategies and powerful problem solving techniques, ***quick solution of the problems could be achieved in a few steps.*

We always *analyze the problem for finding useful patterns and apply suitable methods and strategies to solve the problem in mind.*

We will provide explanations along with solutions.

Let us solve the problems to show you what we mean by the above statements. We urge you to try to solve the problems yourself before going through the solutions.

### Solutions to NCERT class 10 level Trigonometry problemsâ€”Set 5

#### Problem 1.

Prove the identity,

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

#### Solution 1. Problem analysis and useful pattern identification

On initial problem analysis, the lightly hidden pattern of $(1-2\sin^2 \theta)$ in the numerator and $(2cos^2 \theta - 1)$ in the denominator could be identified just by taking out the factors, $\sin \theta$ and $\cos \theta$.

We know these two can be further simplified by using trigonometric relation,

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

Going ahead, the 1's in the two expressions are substituted by $\sin^2 \theta+\cos^2 \theta$ and immediately both the expressions become same,

$\cos^2 \theta - \sin^2 \theta$

The two cancel out leaving $\tan \theta$ as the final result.

Let us show you the deductions.

#### Solution 1: Problem solving execution: Deductive steps

The LHS,

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

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

$=\displaystyle\frac{\sin \theta(\sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta-\sin^2 \theta - \cos^2 \theta)}$, substituting $\sin^2 \theta+\cos^2 \theta$ for 1 in both numerator and denominator,

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

$=\tan \theta$

$=RHS$.

Proved.

This is a problem in which the useful pattern was lightly hidden and the problem could be solved easily and quickly in mind.

Two useful relations to remember,

$1-2\sin^2 \theta=2\cos^2 \theta - 1=\cos^2 \theta - \sin^2 \theta$.

#### Problem 2

Prove the identity,

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

#### Solution 2: Problem analysis, key pattern identification and strategic decision

The key pattern that could be identified in both the squared expressions isâ€”the terms are **inverses to each other**. In such cases, when the *sum of two such terms are squared the middle terms are simplified to just numeric 2*. This is a powerful * property of inverses* that we always use. It is an important problem solving pattern and method.

If you are curious, read our article on, **Principle of interaction of inverses for elegant problem solving.**

So the decision has been to expand the squared expressions. Simplifying, we get the RHS in two steps.

Let us proceed with the deductions.

#### Solution 2: Problem solving steps

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

$=\sin^2 \text{A} + \cos^2 \text{A} + 4 + \text{cosec}^2 \text{A} + \sec^2 \text{A}$, expanding and rearranging the terms,

$=1 + 4 + 2 + (\text{cosec}^2 \text{A} -1) + (\sec^2 \text{A}-1)$

$=7 + \tan^2 \text{A}+ \text{cot}^2 \text{A}$

$=RHS$

Proved.

This is also a problem easily solvable in mind which you should always try. This approach will improve your ability to solve problems in fewer number of steps and hence faster.

#### End note

We consider the approach to the solution more important than the solution itself. It is a way of thinking, not just solution of a problem.

### Further reading materials on Trigonometry

#### NCERT solutions for class 10 maths

**NCERT solutions for class 10 maths Ttrigonometry Set 6**

**NCERT solutions for class 10 maths Trigonometry Set 5**

**NCERT solutions for class 10 maths Trigonometry Set 4**

**NCERT solutions for class 10 maths Trigonometry Set 3**

**NCERT solutions for class 10 maths Trigonometry Set 2**

**NCERT solutions for class 10 maths Trigonometry Set 1**

#### Class 10 Maths

**How to solve school math problems in a few direct steps Trigonometry 5**

**How to solve school math problems in a few steps and many ways Trigonometry 4**

**How to solve school math problems in a few simple steps Trigonometry 3**

**How to solve school math problems in a few simple steps Trigonometry 2**

**How to solve school math problems in a few steps Trigonometry 1**

#### Tutorials, question and answer sets for competitive exams valuable for school level

You may refer to a fair amount of * concise tutorials* and

*created for competitive exams at the page containing list of links,*

**MCQ type question and answer sets**