Question

Prove that: (secA-cosec A) (1+cot A +tan A) =( sec^2A/cosecA)-(Cosec^2A/secA)​

Answer 1

Step-by-step explanation:

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

$=(\sec A - \csc A)\left(1 + \dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A} \right)$

$=(\sec A - \csc A)\left(1 + \dfrac{\cos^2 A + \sin^2 A}{\sin A\cos A} \right)$

$=(\sec A - \csc A)\left(\dfrac{1 + \sin A \cos A}{\sin A \cos A} \right)$

$=\left(\dfrac{\frac{1}{\cos A} - \frac{1}{\sin A}+\sin A - \cos A}{\sin A\cos A}\right)$

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

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

$=\dfrac{\sin^3A - \cos^3A}{\sin^2A\cos^2A}$

$=\dfrac{\sin A}{\cos^2A} - \dfrac{\cos A}{\sin^2A}$

$=\left(\dfrac{1}{\cos A}\right)\left(\dfrac{\sin A}{1}\right) - \left(\dfrac{1}{\sin^2A}\right) \left(\dfrac{\cos A}{1}\right)$

$=\sec^2A\csc A - \csc^2A\sec A$