question_answer

Prove the following identity:

\[{{\left[ \frac{1-\tan \,A}{1-\cot \,A} \right]}^{2}}={{\tan }^{2}}A:\angle A\] is acute

Answer:

Given, \[{{\left[ \frac{1-\tan \,A}{1-\cot \,A} \right]}^{2}}={{\tan }^{2}}A:\angle A\] is acute

\[L.H.S.={{\left[ \frac{1-\tan \,A}{1-\cot \,A} \right]}^{2}}\]

\[={{\left[ \frac{1-\frac{\sin \,A}{\cos \,A}}{1-\frac{\cos \,A}{\sin \,A}} \right]}^{2}}\]

\[={{\left[ \frac{\frac{\cos \,A-\sin \,A}{\cos \,A}}{\frac{\sin \,A-\cos \,A}{\sin \,A}} \right]}^{2}}\]

\[={{\left[ \frac{(cos\,A-sin\,A)sin\,A}{-(cos\,A-sin\,A)cos\,A} \right]}^{2}}\]

\[={{\left[ -\frac{sin\,A}{cos\,A} \right]}^{2}}\]

\[={{[-\,tan\,A]}^{2}}\]

\[={{\tan }^{2}}A=R.H.S.\]              Hence Proved.