Prove the following:
$ \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 (90^{\circ}- \theta)}{\cot \theta} = 2 $

To do:

We have to prove that \( \frac{\cos \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right) \tan \theta}{\operatorname{cosec}\left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)} \) \(+\frac{\tan (90^{\circ}- \theta)}{\cot \theta} = 2 \).

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

$cos\ (90^{\circ}- \theta) = sin\ \theta$

$tan\ (90^{\circ}- \theta) = cot\ \theta$

$cot\ (90^{\circ}- \theta) = tan\ \theta$

$cosec (90^{\circ}- \theta) = sec\ \theta$

$sec\ (90^{\circ}- \theta) = cosec\ \theta$

$sin\ \theta \times cosec\ \theta=1$

$cos\ \theta \times sec\ \theta=1$

Therefore,

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

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

$=1+1$

$=2$

Hence proved.   

Tutorialspoint

Updated on: 10-Oct-2022

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