(solved)Question 6 SSC-CGL 2020 March 4 Shift 1

If secθ - tanθ = x/y, find the value of sinθ?
(Rev. 20-Jan-2024)

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Question 6
SSC-CGL 2020 Mar 4 Shift 1

If secθ - tanθ = x/y, find the value of sinθ?

Solution in Brief

Since sec²θ - tan²θ = 1, we can rationalize LHS of secθ - tanθ = x/y to get secθ + tanθ = y/x. Then add these two to get sec θ = (x² + y²)/(2xy), then get cos θ, and then sin θ as (x² - y²)/(x² + y²) answer.

Solution in Detail

We have been given that

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

Take the reciprocal

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

but $\displaystyle 1 = \sec^2 \theta - \tan^2 \theta$

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

Canceling $\displaystyle \sec \theta - \tan \theta$, we get

$\displaystyle \sec \theta + \tan \theta = \frac yx \text{ . . . (2)}$

Adding (1) and (2)

$\displaystyle \sec \theta = \frac xy + \frac yx$

$\displaystyle \implies \sec \theta = \frac{x^2 + y^2}{2xy}$

$\displaystyle \therefore \cos \theta = \frac{2xy}{x^2 + y^2}$

$\displaystyle \therefore B = 2xy, H = x^2 + y^2$

By pythagoras, $\displaystyle P = x^2 - y^2$

$\displaystyle \therefore \sin \theta = \frac{P}{H} = \frac{x^2 - y^2}{x^2 + y^2}\:\underline{Ans}$

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