## Hard trigonometry questions and solutions for SSC CGL Set 65

Hard trigonometry questions and solutions for competitive exams, SSC CGL 65. Solutions are quick. Basic and advanced concepts and techniques used.

For best results, take the test first before going through the solutions.

Otherwise, you won't be able to appreciate the finer points of the quick and elegant solutions.

For your convenience, the specially selected question set is repeated for practice.

The sections are,

**Hard trigonometry questions for competitive exams SSC CGL 65 (10 selected questions).****Answers to the 10 hard Trigonometry questions for competitive exams SSC CGL 65.****Quick easy to understand solutions for the 10 selected trigonometry questions SSC CGL 65.**

*You may move directly to any of the above sections by clicking its link and return by clicking on browser back button.*

### Hard trigonometry questions for competitive exams SSC CGL Set 65 - time to solve 12 mins

**Problem 1.**

If $2-cos^2 \theta=3sin \theta{cos \theta}$, where $sin \theta \neq cos \theta$, the value of $tan \theta$ is,

- $0$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{1}{3}$

**Problem 2.**

If $sin \theta + cos \theta =\sqrt{2}cos(90^0- \theta)$, then $cot \theta$ is,

- $\sqrt{2}-1$
- $\sqrt{2}+1$
- $0$
- $\sqrt{2}$

**Problem 3.**

If $(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$ then the value of $tan \theta$ is,

- $\displaystyle\frac{1}{2ab}(a^2+b^2)$
- $\displaystyle\frac{1}{2}(a^2-b^2)$
- $\displaystyle\frac{1}{2}(a^2+b^2)$
- $\displaystyle\frac{1}{2ab}(a^2-b^2)$

**Problem 4.**

The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,

- 4
- 0
- 2
- 1

**Problem 5.**

If $cot \theta + cosec \theta =3$, and $\theta$ an acute angle, the value of $cos \theta$ is,

- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{4}$

**Problem 6.**

If $xcos \theta - sin \theta=1$, then the value of $x^2-(1+x^2)sin \theta$ is,

- $1$
- $0$
- $2$
- $-1$

**Problem 7.**

If $\theta=60^0$, then the value of $\displaystyle\frac{1}{2}\sqrt{1+ sin \theta} + \displaystyle\frac{1}{2}\sqrt{1- sin \theta}$ is,

- $cot \displaystyle\frac{\theta}{2}$
- $cos \displaystyle\frac{\theta}{2}$
- $sec \displaystyle\frac{\theta}{2}$
- $sin \displaystyle\frac{\theta}{2}$

**Problem 8.**

If $3sin \theta + 5cos \theta =5$, ($0\lt \theta \lt 90^0$), then the value of $5sin \theta-3cos \theta$ will be,

- 1
- 2
- 5
- 3

**Problem 9.**

If $tan \theta = \displaystyle\frac{1}{\sqrt{11}}$, and $0\lt \theta \lt 90^0$, then the value of $\displaystyle\frac{cosec^2 \theta - sec^2 \theta}{cosec^2 \theta + sec^2 \theta}$ is,

- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{6}{7}$

**Problem 10.**

If $tan^2 \theta=1-e^2$, then the value of $sec \theta + tan^3 \theta{cosec \theta}$ is equal to,

- $(2+e^2)^{\frac{3}{2}}$
- $(2+e^2)^{\frac{1}{2}}$
- $(2-e^2)^{\frac{3}{2}}$
- $(2-e^2)^{\frac{1}{2}}$

The **answers** are given below.

### Answers to the 10 hard trigonometry questions for competitive exams SSC CGL Set 65

**Problem 1.** **Answer:** b: $\displaystyle\frac{1}{2}$.

**Problem 2.** **Answer:** a: $\sqrt{2}-1$.

**Problem 3.** **Answer:** d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$.

**Problem 4.** **Answer:** Option c: 2.

**Problem 5.** **Answer:** c: $\displaystyle\frac{4}{5}$.

**Problem 6.** **Answer:** a: $1$.

**Problem 7.** **Answer:** b: $cos \displaystyle\frac{\theta}{2}$.

**Problem 8.** **Answer:** d: 3.

**Problem 9.** **Answer:** a: $\displaystyle\frac{5}{6}$.

**Problem 10.** **Answer:** c: $(2-e^2)^{\frac{3}{2}}$.

### Hard trigonometry questions and solutions for competitive exam, SSC CGL Set 65 - testing time was 12 mins

**Problem 1.**

If $2-cos^2 \theta=3sin \theta{cos \theta}$, where $sin \theta \neq cos \theta$, the value of $tan \theta$ is,

- $0$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{1}{3}$

**Solution 1: Problem analysis and execution**

As the target is in terms of $tan \theta$, we decide to transform the given expression in terms of $tan \theta$ and then solve for $tan \theta$. To achieve the desired transformation, we need to divide the equation by $cos^2 \theta$. This is target or end state driven input transformation.

The given expression,

$2-cos^2 \theta=3sin \theta{cos \theta}$,

Or, $2sec^2 \theta-1=3tan \theta$,

Or, $2(1+tan^2 \theta)-1=3tan \theta$,

Or, $2tan^2 \theta-3tan \theta+1=0$,

Or, $(2tan \theta-1)(tan \theta -1)=0$,

As $sin \theta \neq cos \theta$, $tan \theta \neq 1$.

So,

$2tan \theta=1$,

Or, $tan \theta =\displaystyle\frac{1}{2}$.

**Answer:** b: $\displaystyle\frac{1}{2}$.

**Key concepts used:** * End state analysis* --

*--*

**Target driven input transformation***--*

**Variable reduction technique,**we have reduced the number of variables or functions, to just 1 in the quadratic equation*.*

**rich trigonometry concepts -- Factorization of quadratic equation -- Basic algebra concepts****Problem 2.**

If $sin \theta + cos \theta =\sqrt{2}cos(90^0- \theta)$, then $cot \theta$ is,

- $\sqrt{2}-1$
- $\sqrt{2}+1$
- $0$
- $\sqrt{2}$

**Solution 2: Problem analysis and execution**

Using * complementary trigonometric functions* concept we transform first the $cos$ function on the RHS to an equivalent $sin$ function,

$sin \theta + cos \theta =\sqrt{2}cos(90^0- \theta)$,

Or, $sin \theta + cos \theta =\sqrt{2}sin \theta$.

Dividing the equation by $sin \theta$ leaves only $cot \theta$ in the equation and so the result in two steps,

$sin \theta + cos \theta =\sqrt{2}sin \theta$,

Or, $1 + cot \theta =\sqrt{2}$,

Or, $cot \theta =\sqrt{2}-1$.

**Answer:** a: $\sqrt{2}-1$.

**Key concepts used:** ** Complementary trigonometric functions** --

**Target driven input transformation.****Problem 3.**

If $(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$ then the value of $tan \theta$ is,

- $\displaystyle\frac{1}{2ab}(a^2+b^2)$
- $\displaystyle\frac{1}{2}(a^2-b^2)$
- $\displaystyle\frac{1}{2}(a^2+b^2)$
- $\displaystyle\frac{1}{2ab}(a^2-b^2)$

**Solution 3: Problem analysis and execution Stage 1: increasing harmony in given expression**

The expression seemed to be **discordant with lack of harmony and too many terms.**

#### Principle of harmony or dicordance in expressions

The variables are the functions here. The first important characteristic is, the two functions involved $sin \theta$ and $cos \theta$ are friendly trigonometric function pair, though in comparison, this pair of functions does not have as much potential in simplification of expressions as the other two friendly trigonometric function pairs, $sec \theta$, $tan \theta$ and $cosec \theta$, $cot \theta$.

Each of these latter two have the * powerful property of mutually inverse expression relationship* of the form, $sec \theta + tan \theta=\displaystyle\frac{1}{sec \theta - tan \theta}$ which in general carries high potential in simplifying complex expressions easily and elegantly.

Coming back to the *discordant form of the given expression*, the discordance or * lack of harmony originates* from the

*$sin \theta$ and $cos \theta$. The function $sin \theta$ is associated with a factor $(a^2-b^2)$ whereas its pair partner $cos \theta$ is associated with a term $2ab$ which is very different in structure and form from $(a^2-b^2)$. If the factor of $cos \theta$ were, say, $(a^2+b^2)$ we could have accepted the expression as an expression in harmony.*

**term association of the friendly functions**The * principle of harmony or discordance in expressions* states,

The more is the harmony or less is the dicordance in an expression, chances of existence of an elegant conceptual solution in a few steps will be that much more.

In absence of no clear elegant solution path visible at this stage, decision is taken to transform the expression to more harmonous form which should reveal elegant solution path after transformation.

If both sides of the equation are divided by $cos \theta$, * thus breaking the discordant association of $cos \theta$* with $2ab$, in a single step we transform the equation in terms of

*of $sec \theta$ and $tan \theta$ and on top of it form term associations of the variables, $sec \theta$ and $tan \theta$ that are similar in structure and form, that is, $(a^2+b^2)$ and $(a^2-b^2)$,*

**more preferred friendly trigonometric function pair**$(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$,

$(a^2-b^2)tan \theta + 2ab=(a^2+b^2)sec \theta$.

In one simple action, the expression is transformed to a much more balanced and harmonious form with lesser discordance.

**Solution 3: Problem Solving Stage 2: further increasing harmony in given expression**

Knowing the power of the mutually inverse expressions, $sec \theta +tan \theta$ and $sec \theta-tan \theta$ of the friendly trigonometric function pair, $sec \theta$, and $tan \theta$, the immediate next step in increasing the harmony in the expression further is to apply the * principle of collection of friendly terms* and bring the terms involving $sec \theta$ and $tan \theta$ together by applying again another powerful rich algebraic

*,*

**technique of collection of friendly terms**From the previous stage,

$(a^2-b^2)tan \theta + 2ab=(a^2+b^2)sec \theta$,

Or, $2ab=a^2(sec \theta - tan \theta) + b^2(sec \theta + tan \theta)$.

#### Solution 3: **Problem Solving Stage 3:** Forming mutually inverse coefficient factors

With RHS transformed into a promising form, we turn our attention to the last offending discordant factor term $ab$ in the LHS.

This term is in no way is similar to the coefficients $a^2$ or $b^2$, but if we divide the equation by $ab$, not only is it removed from the expression, but also a pair of mutually inverse coefficient factors, $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{b}{a}$ is formed in the RHS. This situation is still more welcome as the other two factors of the two terms, $sec \theta - tan \theta$ and $sec \theta + tan \theta=\displaystyle\frac{1}{sec \theta-tan \theta}$ are already present as a **second pair of mutually inverse expressions.**

From the previous stage,

$2ab=a^2(sec \theta - tan \theta) + b^2(sec \theta + tan \theta)$,

Or, $2=\displaystyle\frac{a}{b}(sec \theta-tan \theta)+\displaystyle\frac{b}{a}(sec \theta+tan \theta)$.

#### Solution 3: **Problem Solving Stage 4:** Reduction in number of variables by component expression substitution, friendly trigonometric function pairs

* Reduction in number of variables technique* states,

In simplifying an expression, the more we are able to reduce the number of variables in the expression, easier and simpler would be the steps to the solution.

To reduce number of variables and elements in the expression drastically, we find the situation ideal for applying * component expression substitution* by using an intermediate dummy variable $z=\displaystyle\frac{a}{b}(sec \theta-tan \theta)$, and reduce the number variables to just 1.

From the previous result,

$2=\displaystyle\frac{a}{b}(sec \theta-tan \theta)+\displaystyle\frac{b}{a}(sec \theta+tan \theta)$,

Or, $2=z+\displaystyle\frac{1}{z}$,

where,

$z=\displaystyle\frac{a}{b}(sec \theta-tan \theta)$.

So the given expression is transformed by the substitution to,

$2=z+\displaystyle\frac{1}{z}$,

Or, $z^2-2z+1=0$.

Or, $(z-1)^2=0$,

Or, $z=\displaystyle\frac{a}{b}(sec \theta-tan \theta)=1$, by **Reverse substitution**,

Or, $sec \theta - tan \theta=\displaystyle\frac{b}{a}$.

By friendly trigonometric function pairs concept then,

$sec \theta - tan \theta=\displaystyle\frac{b}{a}$,

Or, $\displaystyle\frac{1}{sec \theta + tan \theta} = \displaystyle\frac{b}{a}$,

Or, $sec \theta + tan \theta = \displaystyle\frac{a}{b}$.

This is familiar grounds and we just substract the value of $sec \theta -tan \theta$ from $sec \theta + tan \theta$ to get $tan \theta$,

$2tan \theta = \displaystyle\frac{a}{b} - \displaystyle\frac{b}{a}$,

Or, $tan \theta =\displaystyle\frac{1}{2ab}(a^2-b^2)$.

**Answer:** d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$.

**Key concepts used:** * Problem analysis* --

*--*

**Key pattern identification***--*

**Principle of harmony or dicordance in expressions -- Discordant variable associations -- Friendly trigonometric function pairs concepts***--*

**Mutually inverse expression resource***--*

**Principle of collection of friendly terms***--*

**Component expression substitution***--*

**Variable reduction technique***--*

**Solving a simpler problem***--*

**Reverse substitution technique***--*

**Basic algebraic concepts***--*

**Rich algebraic techniques***--*

**Minimum order simplest solution principle***--*

**Delayed raise in powers technique***--*

**Efficient simplification**

**Multilevel abstraction -- Degree of abstraction.****Problem 4.**

The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,

- 4
- 0
- 2
- 1

**Solution 4: Problem analysis and execution: **Denominator equalization by denominator rationalization

Though simple in nature, in any problem involving simplification of multiple additive fractional terms, * denominator equalization technique* always achieves an elegant solution. In this case also, we look for an way to equalize the denominators of the two terms.

A second great result we have experienced so often,

If you look for some special result in particular, chances will be high that you will get it.

Conversely,

If you don't look for a new opportunity, you will never be able to discover it.

In this problem, when we searched for an way to equalize the denominators, the lightly hidden opportunity of rationalizing the denominator of the second term revealed itself.

Let us show how.

The given expression,

$E=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta(1-sin \theta)}{1-sin^2 \theta}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta(1-sin \theta)}{cos^2 \theta}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{1-sin \theta}{cos \theta}\right)-2tan^2 \theta$

$=2sec \theta\left(\displaystyle\frac{1}{cos \theta}\right)-2tan^2 \theta$

$=2sec^2 \theta - 2tan^2 \theta$

$=2$.

**Answer:** Option c: 2.

**Key concepts used: Deductive reasoning **to attach priority to denominator equalization as a means to elegant simplification and identification and application of subsequent problem solving steps in a chained sequence

**--**--

*Denominator equalization**--*

**Key patttern identification***--*

**Denominator rationalization***--*

**Basic trigonometry concepts***--*

**Basic algebra concepts***--*

**Rich algebra techniques**

**Efficient simplification.****Problem 5.**

If $cot \theta + cosec \theta =3$, and $\theta$ an acute angle, the value of $cos \theta$ is,

- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{4}$

**Solution 5: Problem analysis and execution: P**rinciple of friendly trigonometric function pair and solving linear expressions

We are already aware of the * principle of friendly trigonometric function pairs*. The results from this rich principle are simple and based on fundamental concepts but have far reaching potential in simplifying trigonometric expressions quickly.

Two of the results of this principle are,

$sec \theta + tan \theta =\displaystyle\frac{1}{sec \theta - tan \theta}$, and

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

In this problem, the given expression directly conforms to this principle and rest of the steps are trivial algebraic simplification in linear expressions.

Given expression,

$cot \theta + cosec \theta =3$,

Or, $\displaystyle\frac{1}{cosec \theta - cot \theta}=3$, only the LHS changes,

Or, $cosec \theta - cot \theta =\displaystyle\frac{1}{3}$.

Adding the two equations,

$2cosec \theta=3+\displaystyle\frac{1}{3}=\displaystyle\frac{10}{3}$, $cot \theta$ is eliminated and we get the most desired single variable value

Or $cosec \theta = \displaystyle\frac{5}{3}$,

Or, $sin \theta =\displaystyle\frac{3}{5}$,

Or, $cos \theta =\sqrt{1-\displaystyle\frac{3^2}{5^2}}=\displaystyle\frac{4}{5}$, as $\theta$ is acute $cos \theta$ can't be negative.

**Answer:** c: $\displaystyle\frac{4}{5}$.

**Key concepts used:** * Key pattern identification* --

*--*

**rich concept of friendly trigonometric function pairs***--*

**basic algebraic concepts**

**solving simple linear equations**

**.****Problem 6.**

If $xcos \theta - sin \theta=1$, then the value of $x^2-(1+x^2)sin \theta$ is,

- $1$
- $0$
- $2$
- $-1$

**Solution 6: **Problem analysis and execution

Recognizing that the target expression involves only $x^2$ and no single power of $x$, before inevitable squaring of $x$, first we transform the given expression to,

$xcos \theta - sin \theta=1$,

Or, $xcos \theta =1+ sin \theta$.

Now when we raise both sides of the equation to the power 2, no term in single power of $x$ is created,

$x^2cos^2 \theta =(1+ sin \theta)^2$.

At this stage in the elegant solution, we detect a possibility to **simplify the expression in trigonometric functions** significantly,

$x^2cos^2 \theta =(1+ sin \theta)^2$,

Or, $x^2(1-sin^2 \theta)=(1+sin \theta)^2$,

Or, $x^2(1-sin \theta)=1 + sin \theta$, $(1+sin \theta)$ factor cancels out from two sides of the equation,

Or, $x^2-(1+x^2)sin \theta=1$.

**Answer:** a: $1$.

**Key concepts used:** * End state analysis* --

*, ensuring that no single power term in $x$ is created after squaring the equation --*

**Target driven input transformation***in detecting the possibility of cancelling out $(1+ sin \theta)$ as a factor --*

**Key pattern identification***.*

**basic trigonometric concepts -- efficient simplification****Problem 7.**

If $\theta=60^0$, then the value of $\displaystyle\frac{1}{2}\sqrt{1+ sin \theta} + \displaystyle\frac{1}{2}\sqrt{1- sin \theta}$ is,

- $cot \displaystyle\frac{\theta}{2}$
- $cos \displaystyle\frac{\theta}{2}$
- $sec \displaystyle\frac{\theta}{2}$
- $sin \displaystyle\frac{\theta}{2}$

**Solution 7: Problem analysis**

Substituting $sin \theta=sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$ in the target expression we will get a square root under another square root in both $\sqrt{1+sin \theta}$ and $\sqrt{1-sin \theta}$.

Unless the surd expression under the square root is transformed into a square of sum of surd, the problem cannot be solved. Thus our main objective is turned into expressing the following two expressions into square of sum of surd expressions,

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

$1-sin \theta=1-\displaystyle\frac{\sqrt{3}}{2}$.

**Solution 7: Problem solving execution**

Let us first transform the problematic surd expressions into square of sum of surd expressions.

$1+sin \theta=1+\displaystyle\frac{\sqrt{3}}{2}$

$=\displaystyle\frac{\sqrt{3}+2}{2}$,

$=\displaystyle\frac{2\sqrt{3}+4}{4}$, multiplying and dividing by 2

$=\displaystyle\frac{(\sqrt{3}+1)^2}{2^2}$.

So,

$\sqrt{1+sin \theta}=\displaystyle\frac{1}{2}(\sqrt{3}+1)$, and

$\sqrt{1-sin \theta}=\displaystyle\frac{1}{2}(\sqrt{3}-1)$,

Thus we have the target expression,

$E=\displaystyle\frac{1}{2}\sqrt{1+ sin \theta} + \displaystyle\frac{1}{2}\sqrt{1- sin \theta}$

$=\displaystyle\frac{1}{4}(\sqrt{3}+1)+\displaystyle\frac{1}{4}(\sqrt{3}-1)$

$=\displaystyle\frac{\sqrt{3}}{2}$

$=cos 30^0$

$=cos \displaystyle\frac{\theta}{2}$.

**Answer:** b: $cos \displaystyle\frac{\theta}{2}$.

**Key concepts used:** * Key pattern identification* --

*--*

**Deductive reasoning***--*

**Conversion to square of sum surd expression by multiplying and dividing the surd expression by 2 to create a factor of 2 with the surd term***--*

**Efficient simplification**

**Basic trigonometry concepts.****Problem 8.**

If $3sin \theta + 5cos \theta =5$, ($0\lt \theta \lt 90^0$), then the value of $5sin \theta-3cos \theta$ will be,

- 1
- 2
- 5
- 3

**Solution 8: Problem analysis and execution**

The similarity between the target and the given expression being coefficient reversal between $sin \theta$ and $cos \theta$, we look for a way to convert the given expression in the form of a sum (addditive or subtractive) of $sec \theta$, $tan \theta$, or, $cosec \theta$, $cot \theta$. We expect to find such a pattern, because it will automatically give us its mutually inverse sum which in turn will reverse the coefficients.

Surely enough we find the subtractive sum of $cosec \theta$ and $cot \theta$. It was kind of inevitable by our reckoning.

Let us show how.

The given expression,

$3sin \theta + 5cos \theta =5$,

Or, $3+5cot \theta=5cosec \theta$, we divide the equation by $sin \theta$, we can do so as, $sin \theta \neq 0$

Or, $5(cosec \theta - cot \theta) = 3$.

Applying principle of friendly trigonometric functions,

$5(cosec \theta - cot \theta) = 3$,

Or, $\displaystyle\frac{5}{cosec \theta +cot \theta}=3$,

Or, $5 =3(cosec \theta + cot \theta)$,

Or, $5sin \theta =3+3cos \theta$, multiplying the equation by $sin \theta$ this time

Or, $5sin \theta -3cos \theta=3$.

**Answer:** d: 3.

**Key concepts used:** * End state analysis* --

*--*

**Key pattern identification***--*

**Coefficient reversal***--*

**Target driven input transformation**

**Friendly trigonometric function pairs concepts.****Problem 9.**

If $tan \theta = \displaystyle\frac{1}{\sqrt{11}}$, and $0\lt \theta \lt 90^0$, then the value of $\displaystyle\frac{cosec^2 \theta - sec^2 \theta}{cosec^2 \theta + sec^2 \theta}$ is,

- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{6}{7}$

**Solution 9: Problem analysis and execution**

Given value being for $tan \theta$ and the value itself being a surd, with choice values all without surds, it points towards transforming the target expression in terms of $tan^2 \theta$. So we take up the task of transforming the target expression in terms of $tan^2 \theta$ in the simplest way,

$E=\displaystyle\frac{cosec^2 \theta - sec^2 \theta}{cosec^2 \theta + sec^2 \theta}$

$=\displaystyle\frac{1 - tan^2 \theta}{1 +tan^2 \theta}$, dividing numerator and denominator by $cosec^2 \theta$,

$=\displaystyle\frac{1 - \displaystyle\frac{1}{11}}{1 +\displaystyle\frac{1}{11}}$

$=\displaystyle\frac{\displaystyle\frac{10}{11}}{\displaystyle\frac{12}{11}}$

$=\displaystyle\frac{10}{12}$

$=\displaystyle\frac{5}{6}$.

**Answer:** a: $\displaystyle\frac{5}{6}$.

**Key concepts used:** **End state analysis -- free resource use -- deductive reasoning -- key pattern identification -- efficient simplification.**

**Problem 10.**

If $tan^2 \theta=1-e^2$, then the value of $sec \theta + tan^3 \theta{cosec \theta}$ is equal to,

- $(2+e^2)^{\frac{3}{2}}$
- $(2+e^2)^{\frac{1}{2}}$
- $(2-e^2)^{\frac{3}{2}}$
- $(2-e^2)^{\frac{1}{2}}$

#### Solution 10: Problem analysis

As $e^2$ occurs in unchanged form in the choice values, by using principle of free resources we conlude the first action to take is to simplify the target expression as much as possible, and substitute the value of $tan^2 \theta$ at the very last stage.

#### Solution 10: Problem solving execution

The target expression,

$E=sec \theta + tan^3 \theta{cosec \theta}$

$=sec \theta + tan^2 \theta{sec \theta}$

$=sec \theta(1+ tan^2 \theta)$

$=(sec^2 \theta)^{\frac{1}{2}}(1+tan^2 \theta)$

$=(1+tan^2 \theta)^{\frac{3}{2}}$

$=(2-e^2)^{\frac{3}{2}}$.

**Answer:** c: $(2-e^2)^{\frac{3}{2}}$.

**Key concepts and techniques used:** * Problem analysis and end state analysis*, decision to simplify first instead of substitution --

*--*

**basic indices concepts***--*

**Basic trigonometry concepts***--*

**Basic algebra concepts***.*

**efficient simplification**

**Note:** You will observe that in many of the Trigonometric problems basic and rich algebraic concepts and techniques are to be used. In fact that is the norm. Algebraic concepts are frequently used for elegant solutions of Trigonometric problems.

### Guided help on Trigonometry in Suresolv

To get the best results out of the extensive range of articles of **tutorials**, **questions** and **solutions** on **Trigonometry **in Suresolv, *follow the guide,*

**Trigonometry in Suresolv for SSC CHSL, SSC CGL, SSC CGL Tier II Other Competitive exams.**

**The guide list of articles is up-to-date.**