# (solved)Question 1 SSC-CGL 2018 June 4 Shift 2

$\displaystyle \bigg(\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta}\bigg)^2 + 1 = ?$
(Rev. 18-Jun-2024)

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### Question 1SSC-CGL 2018 June 4 Shift 2

$\displaystyle \bigg(\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta}\bigg)^2 + 1 = ?$

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

### Solution 1

Try to choose a convenient angle and cycle through the options. This is the first thing that you should try. But do a double check of your calculations.

if we take $\displaystyle \theta = 0 \degree$, the LHS becomes 0 + 1 = 1. Only the (a) option gives 1 at $\displaystyle \theta = 0\degree$. Hence (a) is the answer!

### Solution 2

Try approximations. Can be risky at times, but no harm "trying" if you are low on time. If done carefully, the probability of success is quite good.

We know both sin and cos are less than 1. So $\displaystyle \sin^3 \theta$ and $\displaystyle \cos^3 \theta$ are in any case going to be even more smaller, in fact, so small that we can neglect them in comparison to $\displaystyle \sin \theta$ and $\displaystyle \cos \theta$. The given expression becomes $\displaystyle \bigg(\frac{\sin \theta - \approx 0 }{\approx 0 - \cos \theta}\bigg)^2 + 1$ = $\displaystyle \tan^2 \theta + 1$ = $\displaystyle \sec^2 \theta \:\underline{Ans}$

### Solution 3

The classic trigonometric approach.

\begin{aligned} \text{LHS} &= \bigg(\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta}\bigg)^2 + 1\\ &= \bigg[\frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta (2\cos^2 \theta - 1)}\bigg]^2 + 1\\ &=\bigg(\tan \theta \times \frac{\cos 2\theta}{\cos 2 \theta}\bigg)^2 + 1\\ &= \tan^2\theta + 1\\ &= \sec^2 \theta \:\underline{(a) Ans} \end{aligned}

*EXPLANATION: from class XI trigonometry $\displaystyle (2\cos^2 \theta - 1)$ = $\displaystyle(1 - 2\sin^2 \theta) = \cos 2\theta$