# (solved)Question 13 SSC-CGL 2020 March 4 Shift 1

If 5 sin²θ + 14 cosθ = 13, then (secθ - tanθ)/(secθ + tanθ)=?
(Rev. 18-Jun-2024)

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### Question 13SSC-CGL 2020 Mar 4 Shift 1

If 5 sin²θ + 14 cosθ = 13, then (secθ - tanθ)/(secθ + tanθ)=?

### Solution in Detail

To find $\displaystyle \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta}$

Convert sec and tan to sin, cos and simplify the requirement to

$\displaystyle \frac{1 - \sin \theta}{1 + \sin \theta}\text{ . . . (1)}$

Given $\displaystyle 5\sin^2 \theta + 14 \cos \theta = 13$

Use $\displaystyle \sin^2 \theta = 1 - \cos^2 \theta$ to get

$\displaystyle 5\cos^2\theta - 14\cos\theta + 8 = 0$

$\displaystyle (5\cos\theta - 4)(\cos \theta - 2) = 0$

$\displaystyle \therefore \cos \theta = 2, 4/5$

$\displaystyle \cos \theta$ cannot exceed 1,

$\displaystyle \therefore \cos \theta = 4/5$

Pyth triplets are 4, 5 and 3

$\displaystyle \therefore \sin \theta = 3/5$

Putting in (1) and simplifying,

$\displaystyle = \frac{1 - 3/5}{1 + 3/5} = \frac14\:\underline{Ans}$

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