Question

If $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$ then $\tan \theta$ is equal to

• A. $\frac{a^2+b^2}{2ab}$
• B. $\frac{2ab}{a^2-b^2}$
• C. $\frac{a^2-b^2}{2ab}$
• D. $\frac{ab}{a^2-b^2}$

Answer 1

The correct option is C $\frac{a^2-b^2}{2ab}$

$sin\theta =\frac{a^2-b^2}{a^2+b^2}$

$tan\theta =\frac{a^2-b^2}{\sqrt{(a^2+b^2{)}^2-(a^2-b^2{)}^2}}$

$=\frac{a^2-b^2}{\sqrt{a^4+b^4+2a^2{b}^2-(a^2+b^2-2a^2{b}^2)}}$

$tan\theta \Rightarrow \frac{a^2-b^2}{\sqrt{4a^2{b}^2}}=\frac{a^2-b^2}{2ab}$ option C is correct.