Question

tan theta/1-cot theta + cot theta/1-tan theta =1+sec theta cosec theta

Answer 1

Answer:

Step-by-step explanation:

L.H.S.

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\\\\\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\\\\\frac{\sin ^{2}A}{\cos A(\sin A-\cos A)}+\frac{\cos  ^{2}A}{\sin A(\cos A-\sin A)}\\\\\frac{\sin ^{2}A-\cos ^{2}A}{\cos A\sin A(\sin A-\cos A)}\\\\\frac{1+\sin A\cos A}{\cos A\sin A} \\\\1+\sec A\csc A$

=R.H.S.

hence proved.

Answer 2

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

$=\frac{\tan ^{2} \theta}{\tan \theta-1}-\frac{1}{\tan \theta(\tan \theta-1)}$

$=\frac{1}{\tan \theta-1}\left\{\tan ^{2} \theta-\frac{1}{\tan \theta}\right\}$

$=\frac{1}{\tan \theta-1}\left(\frac{\tan ^{3} \theta-1}{\tan \theta}\right)$

$\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

$=\frac{\left\{(\tan \theta-1)\left(\tan ^{2} \theta+1+\tan \theta\right)\right\}}{\{(\tan \theta-1)(\tan \theta)\}}$

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

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

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

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

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\csc \theta \cdot \sec \theta+1$"