If 1/{\sec\\theta-1}+1/{\sec\\theta+1}=2\cosec\\theta, 0^\circ

SSC CGL Model Paper 11

Question : 75 of 100

Marks: +1, -0

1/{\sec\θ-1}+1/{\sec\θ+1}=2\cosec\θ,

0° < \θ < 90°

, then the value of

\sec^2\θ+\tan\θ

Solution:

{\sec\θ+1+\sec\θ-1}/{\sec^2\θ-1}=2 /{\sin\θ}

{2\sec\θ}/{\tan^2\θ}=2/{\sin\θ}

2/{\cos\θ}×{\cos^2\θ}/{\sin^2\θ}=2/{\sin\θ}

2\cot\θ=2

\cot\θ=1

\θ=45°

∴\sec^2\θ+\tan\θ=2+1=3

Go to Question:

Prev Question

Next Question