If \sec\\theta=\frac{k^2+1}{k^2-1},0^\circ

SSC CGL Model Paper 13

Question : 55 of 100

Marks: +1, -0

\sec\θ={k^2+1}/{k^2-1},0° <\θ° <90°,

then the value of

\cosec\θ +\cos\θ ?

Solution:

\sec\θ =H/B={k^2+1}/{k^2-1}

P=2K

Now,

\cosec\θ +\cot\θ

⇒H/B+B/P⇒{2k^2}/{2k}=k

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