Question

If $\sin$ $\theta =\frac{a^2-b^2}{a^2+b^2}$, then Find other Ratios

Answer 1

We have,

sin A = $\frac{Perpendicular}{Hypotenuse}=\frac{a^2-b^2}{a^2+b^2}$

So, we draw a right triangle right angled at B such that

Perpendicular = $a^2-b^2$ and, Hypotenuse = $a^2+b^2$. and $\angle BAC=\theta$

By Pythagoras theorem, we have

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow A{B}^2=(a^2+b^2{)}^2-(a^2-b^2{)}^2$

$\Rightarrow A{B}^2=(a^4+b^4+2a^2{b}^2)-(a^4+b^4-2a^2{b}^2)$

$\Rightarrow A{B}^2=4a^2{b}^2=(2ab{)}^2$

$\Rightarrow AB=2ab$

When we consider the trigonometric ratios of $\angle BAC=\theta$ , we have

Base = AB = 2ab, Perpendicular = BC = $a^2-b^2$, and Hypotenuse = AC = $a^2+b^2$

$\therefore cos\theta =\frac{Base}{Hypotenuse}=\frac{2ab}{a^2+b^2}$

$\Rightarrow tan\theta =\frac{Perpendicular}{Base}=\frac{a^2-b^2}{2ab}$

$\Rightarrow cosec\theta =\frac{Hypotenuse}{Perpendicular}=\frac{a^2+b^2}{a^2-b^2}$

$\Rightarrow sec\theta =\frac{Hypotenuse}{Base}=\frac{a^2+b^2}{2ab}$

and,

$cot\theta =\frac{Base}{Perpendicular}=\frac{2ab}{a^2-b^2}$