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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.