# Solve this

Question:

If $\sin \theta=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$, find the values of all T-ratios of $\theta$.

Solution:

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

As,

$\cos ^{2} \theta=1-\sin ^{2} \theta$

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

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

$=\frac{\left(a^{2}+b^{2}\right)^{2}-\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}$

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

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

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

$\Rightarrow \cos ^{2} \theta=\frac{4 a^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}$

$\Rightarrow \cos \theta=\sqrt{\frac{4 a^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}}$

$\Rightarrow \cos \theta=\frac{2 a b}{\left(a^{2}+b^{2}\right)}$

Also,

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

$=\frac{\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)}{\left(\frac{2 a b}{a^{2}+b^{2}}\right)}$

$=\frac{a^{2}-b^{2}}{2 a b}$

Now,

$\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

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

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

Also,

$\sec \theta=\frac{1}{\cos \theta}$

$=\frac{1}{\left(\frac{2 a b}{a^{2}+b^{2}}\right)}$

$=\frac{a^{2}+b^{2}}{2 a b}$

And,

$\cot \theta=\frac{1}{\tan \theta}$

$=\frac{1}{\left(\frac{a^{2}-b^{2}}{2 a b}\right)}$

$=\frac{2 a b}{a^{2}-b^{2}}$