Prove the following trigonometric identities:$ \frac{\cos ^{2} \theta}{\sin \theta}-\operatorname{cosec} \theta+\sin \theta=0 $

To do:

We have to prove that \( \frac{\cos ^{2} \theta}{\sin \theta}-\operatorname{cosec} \theta+\sin \theta=0 \).

Solution:

We know that,

$\sin \theta\times\operatorname{cosec} \theta=1$.....(i)

$\cos ^{2} \theta+\sin^2 \theta=1$.......(ii)

Therefore,

$\frac{\cos ^{2} \theta}{\sin \theta}-\operatorname{cosec} \theta+\sin \theta=\frac{\cos ^{2} \theta-\operatorname{cosec} \theta\sin \theta+\sin^2 \theta}{\sin \theta}$

$=\frac{\cos ^{2} \theta+\sin^2 \theta-1}{\sin \theta}$                       [From (i)]

$=\frac{1-1}{\sin \theta}$             [From (ii)]                

$=0$              

Hence proved.  

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Updated on: 10-Oct-2022

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