tanθ/secθ-1 + tanθ/secθ+1 is equal to

Question

\(\frac{tanθ}{secθ-1} +\frac{tanθ}{secθ+1}\) is equal to 

A. 2 tan θ 

B. 2 sec θ 

C. 2 cosec θ 

D. 2 tan θ sec θ

Answer 1

To find: \(\frac{tanθ}{secθ-1} +\frac{tanθ}{secθ+1}\) 

Consider \(\frac{tanθ}{secθ-1} +\frac{tanθ}{secθ+1}\) 

∵ tanθ = \(\frac{sinθ}{cosθ}\) ,   secθ = \(\frac{1}{cosθ}\)  

=\(\frac{2sinθ}{sin^2θ}\)  [∵ sin²θ = 1 – cos²θ]

= \(\frac{2}{sinθ}\) 

= 2cosecθ \(\Big[cosecθ = \frac{1}{sinθ}\Big]\)