Question

Tan A-CotA÷Tan A + cot A​

Answer 1

Answer:

$2\sin^2 A - 1$

Step-by-step explanation:

$\dfrac{\tan A - \cot A}{\tan A + \cot A​} =$

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

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

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

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

$= \sin^2 A - \cos^2 A$

Since

$\sin^2 A + \cos^2 A = 1$

we know

$\cos^2 A = 1 - \sin^2 A$

we now substitute to get

$= \sin^2 A - (1 - \sin^2 A)$

$= \sin^2 A - 1 + \sin^2 A)$

$= 2\sin^2 A - 1$