# (solved)Question 15 SSC-CGL 2020 March 4 Shift 3

If secθ + tanθ = p, then the value of (p² - 1)/(p² + 1)
(Rev. 24-Aug-2023)

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### Question 15SSC-CGL 2020 Mar 4 Shift 3

If secθ + tanθ = p, then the value of (p² - 1)/(p² + 1)

### Solution in brief

We can obtain 1/p = secθ - tanθ. Which gives p + 1/p = 2secθ, and p - 1/p = 2tanθ. Hence (p² - 1)/(p² + 1) = (p - 1/p)/(p + 1/p) becomes 2tanθ/2secθ = sinθ ans!

### Solution in detail

Game Plan: Observe that by dividing num and den by p, we can reduce (p² - 1)/(p² + 1) to (p - 1/p)/(p + 1/p). So we can start by finding 1/p.

Given $\displaystyle p = \sec \theta + \tan \theta \text{. . . (1)}$

$\displaystyle \therefore \frac 1p = \frac{1}{\sec \theta + \tan \theta}$

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

$\displaystyle = \frac{\sec \theta - \tan \theta}{\sec^2 \theta - \tan^2 \theta}$

$\displaystyle \therefore \frac 1p = \sec \theta - \tan \theta \text{. . . (2)}$

Subtracting (1) - (2)

$\displaystyle p - \frac 1p = 2\tan \theta$

$\displaystyle p + \frac 1p = 2\sec \theta$
$\displaystyle \therefore \frac{p^2 - 1}{p^2 + 1} = \frac{(p^2 - 1)/p}{(p^2 + 1)/p}$
$\displaystyle = \frac{p - 1/p}{p + 1/p}$
$\displaystyle = \frac{2 \tan \theta}{2 \sec \theta} = \sin \theta \:\underline{Ans}$