Prove the following trigonometric identities:$ \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta $

To do:

We have to prove that \( \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta \).

Solution:

We know that,

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

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

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

Therefore,

LHS

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

$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$

$=\frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta-\sin \theta}$

$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}$

$=\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}$

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

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

R.H.S.

$=1+\tan \theta+\cot \theta$

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

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

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

Here,

LHS $=$ RHS

Hence proved.        

Tutorialspoint

Updated on: 10-Oct-2022

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