### Question 15

SSC-CGL 2018 June 4 Shift 1

If $\displaystyle \sin \theta = \frac{p^2 - 1}{p^2 + 1}$, then $\displaystyle \cos \theta = ?$

- $\displaystyle\frac{2p}{p^2 - 1}$
- $\displaystyle\frac{2p}{p^2 + 1}$
- $\displaystyle\frac{p}{p^2 + 1}$
- $\displaystyle\frac{p}{p^2 - 1}$

### Solution 1

Take any convenient value of $\displaystyle p$, say, $\displaystyle p = 1$

$\displaystyle \implies \sin \theta = 0$, hence $\displaystyle \cos \theta$ should be $\displaystyle 1$.

Cycling through the options with $\displaystyle p = 1$, we immediately get $\displaystyle\frac{2p}{p^2 + 1}$ as the answer!

### Solution 2

REMEMBER: $\displaystyle p^2 - 1, 2p \text{ and } p^2 + 1$ are pythagorean triplets.Given $\displaystyle \sin \theta = \frac{p^2 - 1}{p^2 + 1}$, we can think of a right angled triangle with perpendicular $\displaystyle p^2 - 1$ and hypotenuse as $\displaystyle p^2 + 1$.

The base of the triangle should be $\displaystyle 2p$, giving $\displaystyle \cos \theta = \frac{2p}{p^2 + 1}$

Hence (2) is the answer!

### Solution 3

Without the knowledge of the above rule of triplets.Given $\displaystyle \sin \theta = \frac{p^2 - 1}{p^2 + 1}$, we can think of a right angled triangle with perpendicular $\displaystyle p^2 - 1$ and hypotenuse as $\displaystyle p^2 + 1$.

Cycling the options, we can rule out (1) and (4) because of their denominators. The base of the triangle from option (2) is $\displaystyle 2p$, whereas from option (3) it is $\displaystyle p$.

Testing option (2), $\displaystyle (2p)^2 + (p^2 - 1)^2$ simplifies to $\displaystyle (p^2 + 1)^2$, which verifies pythagoras. Hence (2) is the answer!

### Solution 4

We know $\displaystyle \cos^2 \theta + \sin^2 \theta = 1$

Hence, $\displaystyle \cos^2 \theta = 1 - \bigg(\frac{p^2 - 1}{p^2 + 1}\bigg)^2$

Which simplifies to $\displaystyle \cos^2 \theta = \bigg(\frac{2p}{p^2 + 1}\bigg)^2$

Which gives (2) as the answer!

### Solution 5

The sum of the three sides of a right triangle is always an even numberFrom $\displaystyle \sin \theta = \frac{p^2 - 1}{p^2 + 1}$, we get two sides as $\displaystyle (p^2 - 1)$ and $\displaystyle (p^2 + 1)$, their sum is $\displaystyle 2p^2$, which is even. The third side must be even. From the given options, we can deduce on the basis of denominators that the third side is either $\displaystyle p$ or $\displaystyle 2p$, out of which the latter is guaranteed to be even, and, therefore, the answer!

### Additional Discussion

The options that have $\displaystyle p^2 - 1$ as the denominator cause the expression to become invalid at $\displaystyle p = \pm1$. But $\displaystyle \cos \theta$ is always between $\displaystyle \pm1$, and never undefined. So we could have ruled out those options without any further thinking.

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