Question

The expression $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as

• A. $sinAcosA+1$
• B. $secAcos$ $ecA+1$
• C. $tanA+cotA$
• D. $secA+cos$ $ecA$

Answer 1

The correct option is A $sinAcosA+1$

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

$=\frac{{\sin }^{2}A}{\cos A(\sin A-\cos A)}-\frac{{\cos }^{2}A}{\sin A(\sin A-\cos A)}$

$=\frac{1}{\sin A-\cos A}(\frac{{\sin }^{2}A}{\cos A}-\frac{{\cos }^{2}A}{\sin A})$

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

$=\frac{(\sin A-\cos A)({\sin }^{2}A+{\cos }^{2}A+\sin A\cos A)}{(\sin A-\cos A)\sin A\cos A}$ $(\because a^3-b^3=(a-b\left)\right(a^2+b^2+ab\left)\right)$

$=\frac{1+\sin A\cos A}{\sin A\cos A}=1+\text{sec}A\text{cosec}A$