If \[ \sin x=\frac{a^2-b^2}{a^2+b^2} \] Find the Values of \[ \tan x,\ \sec x \text{ and } \cosec x \]

Solution:

\[ \sin x=\frac{a^2-b^2}{a^2+b^2} \]

Using \[ \sin^2 x+\cos^2 x=1 \]

\[ \cos x = \sqrt{1-\sin^2 x} \]

\[ = \sqrt{ 1- \left( \frac{a^2-b^2}{a^2+b^2} \right)^2 } \]

\[ = \sqrt{ \frac{(a^2+b^2)^2-(a^2-b^2)^2} {(a^2+b^2)^2} } \]

\[ = \sqrt{ \frac{4a^2b^2} {(a^2+b^2)^2} } \]

\[ = \frac{2ab}{a^2+b^2} \]

Therefore,

\[ \tan x = \frac{\sin x}{\cos x} = \frac{a^2-b^2}{2ab} \]

\[ \sec x = \frac{1}{\cos x} = \frac{a^2+b^2}{2ab} \]

\[ \cosec x = \frac{1}{\sin x} = \frac{a^2+b^2}{a^2-b^2} \]