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

If 5 sin²θ + 14 cosθ = 13, then (secθ - tanθ)/(secθ + tanθ)=?
(Rev. 01-May-2025)

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Question 13
SSC-CGL 2020 Mar 4 Shift 1

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

Solution in Detail

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

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

1sinθ1+sinθ . . . (1)\displaystyle \frac{1 - \sin \theta}{1 + \sin \theta}\text{ . . . (1)}

Given 5sin2θ+14cosθ=13\displaystyle 5\sin^2 \theta + 14 \cos \theta = 13

Use sin2θ=1cos2θ\displaystyle \sin^2 \theta = 1 - \cos^2 \theta to get

5cos2θ14cosθ+8=0\displaystyle 5\cos^2\theta - 14\cos\theta + 8 = 0

(5cosθ4)(cosθ2)=0\displaystyle (5\cos\theta - 4)(\cos \theta - 2) = 0

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

cosθ\displaystyle \cos \theta cannot exceed 1,

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

Pyth triplets are 4, 5 and 3

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

Putting in (1) and simplifying,

=13/51+3/5=14Ans\displaystyle = \frac{1 - 3/5}{1 + 3/5} = \frac14\:\underline{Ans}

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