Question

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

Answer 1

\[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\]
We know: 
\[ \sin^2 x + \cos^2 x = 1\]
\[ \cos^2 x = 1 - \sin^2 x\]
\[ = 1 - \left( \frac{a^2 - b^2}{a^2 + b^2} \right)^2 \]
\[ = \frac{\left( a^4 + b^4 + 2 a^2 b^2 \right) - \left( a^4 + b^4 - 2 a^2 b^2 \right)}{\left( a^2 + b^2 \right)^2}\]
\[ = \frac{4 a^2 b^2}{\left( a^2 + b^2 \right)^2}\]
\[ \Rightarrow \cos x = \frac{2ab}{\left( a^2 + b^2 \right)}\]
\[\tan x = \frac{\sin x}{\cos x} = \frac{\frac{a^2 - b^2}{a^2 + b^2}}{\frac{2ab}{a^2 + b^2}} = \frac{a^2 - b^2}{2ab}\]
\[\sec x = \frac{1}{\cos x} = \frac{a^2 + b^2}{2ab}\]
\[cosec x = \frac{1}{\sin x} = \frac{a^2 + b^2}{a^2 - b^2}\]