If θ - θ/θ + θ = 3/5, then the value of θ + θ/θ - θ is:

Correct Answer is : 31+81531 + 8 {15}31+815

Correct Answer is : 31+81531 + 8 {15}31+815

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SSC CGL 3 Mar 2020 Shift 2 Paper

Question : 52 of 75

Marks: +1, -0

\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta} = \frac{3}{5}

\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta}

Solution: 👈: Video Solution

Componendo and dividendo rule:

\frac{a}{b} = \frac{c}{d}

Then

\frac{a+b}{a-b} = \frac{c+d}{c-d}

\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta} = \frac{3}{5}

\Rightarrow \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = \frac{5}{3}

Using Componendo or Dividendo

\frac{\sec\theta}{\tan\theta} = \frac{5+3}{5-3}

\Rightarrow \frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{8}{2}

\Rightarrow \frac{1}{\sin\theta} = 4

\Rightarrow \sin\theta = \frac{1}{4}

Perpendicular

= P

, Hypotenuse

= H

and Base

= B

\sin\theta = \frac{P}{H} = \frac{1}{4}

P = 1

and

H = 4

As we know,

H^2 = P^2 + B^2

\Rightarrow 4^2 = 1^2 + B^2

\Rightarrow B^2 = 16 - 1 = 15

\Rightarrow B = \sqrt{15}

\csc\theta = \frac{H}{P} = 4

\cot\theta = \frac{B}{P} = \sqrt{15}

Now,

\frac{\csc\theta + \cot\theta}{\csc\theta - \cot\theta}

\frac{4+\sqrt{15}}{4-\sqrt{15}}

= \frac{(4+\sqrt{15}) \times (4+\sqrt{15})}{(4-\sqrt{15}) \times (4+\sqrt{15})}

= \frac{(4+\sqrt{15})^2}{16-15}

= (4+\sqrt{15})^2

= 16 + 15 + 2 \times 4 \sqrt{15}

= 31 + 8 \sqrt{15}

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