# (solved)Question 3 SSC-CGL 2018 June 4 Shift 1

Trigonometric: $\displaystyle \frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta} = ?$
(Rev. 03-Aug-2022)

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### Question 3SSC-CGL 2018 June 4 Shift 1

$\displaystyle \frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta} = ?$

1. $\displaystyle \sec \theta \cosec \theta$
2. $\cot \theta$
3. $\tan \theta$
4. $\cos \theta \sin \theta$

### Solution 1

this is a very useful technique for solving complex looking trigonometric equations, also called identities.

Choose a convenient angle such that all four options give different values.

So take $\displaystyle \theta = 30 \degree$

Calculate the options:

A is: $\displaystyle \sec 30 \degree \times \cosec 30 \degree = \frac{4}{\sqrt 3}$

others: $\displaystyle \sqrt 3, \frac{1}{\sqrt 3}$ and $\displaystyle \frac{\sqrt 3}{4}$

Now calculate the expression:

$\displaystyle \therefore \frac{2 + \tan^2 30 \degree + \cot^2 30 \degree}{\sec 30 \degree \cosec 30 \degree}$

$\displaystyle = \frac{2 + \frac13 + 3}{\frac{2}{\sqrt 3} \times 2} = \frac{4}{\sqrt 3}$

Hence the first option is correct.

### Solution 2

Cycle through the options. Try to see if LHS becomes equal to RHS at some stage.

Take the first option.

$\displaystyle \frac{2 + \tan^2 \theta + \cot^2 \theta}{\sec \theta \cosec \theta} = {\sec \theta \cosec \theta}$

Bring $\sec \theta \cosec \theta$ to the denominator of LHS

$\displaystyle \implies \frac{1 + \tan^2 \theta + 1 + \cot^2 \theta}{\sec^2 \theta \cosec^2 \theta} = 1$

$\displaystyle \implies \frac{\sec^2 \theta + \cosec^2 \theta}{\sec^2 \theta \cosec^2 \theta} = 1$

$\displaystyle \implies \sin^2 \theta + \cos^2 \theta = 1$ verified. Hence first option is correct.