Question

Show that:

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

Answer 1

$(1+\cot A+\tan A)(\sin A-\cos A)$

$=\sin A+\sin A\cot A+\sin A\tan A-\cos A-\cos A\cot A-\cos A\tan A$

$=\sin A+\sin A\times \frac{\cos A}{\sin A}+\sin A\times \frac{\sin A}{\cos A}-\cos A-\cos A\times \frac{\cos A}{\sin A}-\cos A\times \frac{\sin A}{\cos A}$

$=\sin A+\cos A+\sin A\tan A-\cos A-\cot A\cos A-\sin A$

$=\sin A\tan A-\cot A\cos A$ ....... $\left(1\right)$

Now, $\sin A\tan A-\cot A\cos A$

$=\sin A\times \frac{\sin A}{\cos A}-\frac{\cos A}{\sin A}\times \cos A$

$={\sin }^{2}A\sec A-{\cos }^{2}A\text{cosec}A$

$=\frac{1}{{\text{cosec}}^{2}A}\times \sec A-\frac{1}{{\sec }^{2}A}\times \text{cosec}A$

$=\frac{\sec A}{{\text{cosec}}^{2}A}-\frac{\text{cosec}A}{{\sec }^{2}A}$ ...... $\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

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