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

Question 15
SSC-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

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)

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

Adding (1) + (2)

$\displaystyle p + \frac 1p = 2\sec \theta$

$\displaystyle \therefore \frac{p^2 - 1}{p^2 + 1} = \frac{(p^2 - 1)/p}{(p^2 + 1)/p}$

$\displaystyle = \frac{p - 1/p}{p + 1/p}$

$\displaystyle = \frac{2 \tan \theta}{2 \sec \theta} = \sin \theta \:\underline{Ans}$