Prove that:$ 1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=\operatorname{cosec} \theta $

To do:

We have to prove that \( 1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=\operatorname{cosec} \theta \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=1+\frac{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}}{1+\frac{1}{\sin \theta}}$

$=1+\frac{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}}{\frac{\sin \theta+1}{\sin \theta}}$

$=1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \frac{\sin \theta}{\sin \theta+1}$

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

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

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

$=\frac{1+\sin \theta}{\sin \theta(\sin \theta+1)}$

$=\frac{1}{\sin \theta}$

$=\operatorname{cosec} \theta$

Hence proved.      

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Simply Easy Learning

Updated on: 10-Oct-2022

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