Question 3 SSC CGL 2018 June 4 Shift 1

Posted by Parveen(Hoven),
Aptitude Trainer and Software Developer

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Question 3 SSC-CGL 2018 June 4
[NOTE: only math questions solved]

Question: $\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. $\cos \theta \sin \theta$

Method 1

Choose an angle at which all four options will give different values. So take $\displaystyle \theta = 30 \degree$

The given expression is $\displaystyle \frac{2 + \tan^2 30 \degree + \cot^2 30 \degree}{\sec 30 \degree \cosec 30 \degree} = \frac{2 + \frac13 + 3}{\frac{2}{\sqrt 3} \times 2} = \frac{4}{\sqrt 3}$

First option will be: $\displaystyle \sec 30 \degree \times \cosec 30 \degree = \frac{4}{\sqrt 3}$

Similarly, the others will be $\displaystyle \sqrt 3, \frac{1}{\sqrt 3}$ and $\displaystyle \frac{\sqrt 3}{4}$

The first option is correct.

Method 2

Cycle through the options and verify as an identity.

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

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

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

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