$\frac{{cosec} \theta}{{cosec} \theta-1}+\frac{{cosec} \theta}{{cosec} \theta+1}-\tan ^2 \theta, 0^{\circ}<\theta<90^{\circ},$ is equal to:
Answer & explanation
Correct answer: option 1
We are given,
\(\frac{cosecθ}{cosecθ-1 }\) + \(\frac{cosecθ}{cosecθ+1 }\) - tan²θ
= \(\frac{cosecθ(cosecθ+1) +cosecθ(cosecθ-1) }{cosec² θ - 1² }\) - tan²θ
= \(\frac{ 2 cosec² θ }{cosec² θ - 1² }\) - tan²θ
{ using , cosec²θ - cot ²θ = 1 }
= \(\frac{ 2 cosec² θ }{cot ² θ}\) - tan²θ
= \(\frac{ 2 }{cos ² θ}\) - tan²θ
= 2 sec²θ - tan²θ
= sec²θ + sec²θ - tan² θ
{ using , sec²θ - tan² θ = 1 }
= sec²θ + 1