Question

Prove that:

$\left(\sec A-\csc A\right)(1+\tan A+\cot A)=\csc A{\sec }^{2}A-\cot A\csc A.$

Answer 1

Consider the given equation.

$(\sec A-\csc A)(1+\tan A+\cot A)=\tan A\sec A-\cot A\csc A$

Consider the LHS.

$=\sec A+\tan A\sec A+\sec A\cot A-\csc A-\csc A\tan A-\csc A\cot A$

$=\sec A+\csc A{\sec }^{2}A+\csc A-\csc A-\sec A-{\csc }^{2}A\cos A$

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

$=\left(\frac{\csc A}{\cos A}\right)\sec A-\left(\frac{\cos A}{\sin A}\right)\csc A$

$=\csc A{\sec }^{2}A-\cot A\csc A$

Therefore,

$LHS=RHS$