The value of cos (90°−θ) sec (90°−θ) tan θcosec (90°−θ) sin (90°−θ) cot (90°−θ)+tan (90°−θ)cot θ is
Question:

The value of $\frac{\cos \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right) \tan \theta}{\operatorname{cosec}\left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}$ is

(a) 1
(b) − 1
(c) 2
(d) −2

Solution:

We have to find: $\frac{\cos \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right) \tan \theta}{\operatorname{cosec}\left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}$

So

$\frac{\cos \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right) \tan \theta}{\operatorname{cosec}\left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}$

$=\frac{\sin \theta \operatorname{cosec} \theta \tan \theta}{\sec \theta \cos \theta \tan \theta}+\frac{\cot \theta}{\cot \theta}$

$=\frac{1 \times \tan \theta}{1 \times \tan \theta}+\frac{\cot \theta}{\cot \theta}$

$=1+1$

$=2$

Hence the correct option is $(c)$

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