Question

If $\sin \theta =\frac{m^2-n^2}{m^2+n^2}$, then what is the value of $\tan \theta$?

• A. $\frac{m^2+n^2}{m^2-n^2}$
• B. $\frac{2m}{m^2+n^2}$
• C. $\frac{m^2-n^2}{2mn}$
• D. $\frac{m^2+n^2}{2mn}$

Answer 1

The correct option is C $\frac{m^2-n^2}{2mn}$

R.E.F image

$\sin \theta =\frac{m^2-n^2}{m^2+n^2}=\frac{P}{h}$

$tan\theta =\frac{P}{B}=\frac{P}{\sqrt{h^2-P^2}}=\frac{m^2-n^2}{\sqrt{(m^2+n^2{)}^2-(m^2-n^2{)}^2}}$

$\Rightarrow tan\theta =\frac{m^2-n^2}{\sqrt{4m^2{n}^2}}=\frac{m^2-n^2}{2mn}$