Question

The expression $\sin A.\cos A(\frac{{\tan }^{2}A}{\tan A-1}+\frac{{\cot }^{2}A}{\cot A-1})$ can be written as:

• A. $\cos A\text{cosec}A+1$
• B. $\tan A+\sec A$
• C. $\sec A+\text{cosec}A$
• D. $\sin A\cos A+1$

Answer 1

The correct option is D $\sin A\cos A+1$

$\frac{{\tan }^{2}A}{\tan A-1}+\frac{{\cot }^{2}A}{\cot A-1}$

$=\frac{\tan A}{\frac{(\tan A-1)}{\tan A}}+\frac{\cot A}{\frac{(\cot A-1)}{\cot A}}$

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

$=\frac{{\tan }^{2}A}{\tan A-1}-\frac{\cot A}{\tan A-1}$

$=\frac{{\tan }^{2}A-\cot A}{\tan A-1}$

$=\frac{{\tan }^{3}A-1}{{\tan }^{2}A-\tan A}$

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

$=\frac{{\sec }^{2}A+\tan A}{\tan A}$

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

$\therefore \sin A\cdot \cos A(\frac{{\tan }^{2}A}{\tan A-1}+\frac{{\cot }^{2}A}{\cot A-1})=1+\sin A\cos A$