Question

If $\sin \alpha =\frac{a^2-b^2}{a^2+b^2}$ then prove that $\cot \alpha =\frac{2ab}{a^2-b^2}$.

Answer 1

Given,

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

or, $cosec\alpha =\frac{a^2+b^2}{a^2-b^2}$.......(1).

We know $cose{c}^2\alpha -{\cot }^2\alpha =1$

or, ${\cot }^2\alpha =cose{c}^2\alpha -1$

or, ${\cot }^2\alpha ={\left(\frac{a^2+b^2}{a^2-b^2}\right)}^2-1$ [ Using (1)]

or, ${\cot }^2\alpha =\frac{(a^2+b^2{)}^2-(a^2-b^2{)}^2}{(a^2-b^2{)}^2}$

or, ${\cot }^2\alpha =\frac{4a^2{b}^2}{(a^2-b^2{)}^2}$

or, $\cot \alpha =\frac{2ab}{a^2-b^2}$.