Question

If sin θ = (a2 - b2) / (a2 + b2), then find all the trigonometric ratios.

Answer 1

I think this will help you

Answer 2

Answer:  The values of all the trigonometric ratios are found as follows :

Step-by-step explanation:  We are given the following trigonometric ratio :

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

We are to find all the other trigonometric ratios.

We will be using the following relations among trigonometric ratios :

$\cos\theta=\pm\sqrt{1-\sin^2\theta},\\\\\tan\theta=\dfrac{\sin\theta}{\cos\theta},\\\\\csc\theta=\dfrac{1}{\sin\theta},\\\\\sec\theta=\dfrac{1}{\cos\theta},\\\\\cot\theta=\dfrac{1}{\tan\theta}.$

We have

$\cos\theta\\\\=\pm\sqrt{1-\sin^2\theta}\\\\=\pm\sqrt{1-\left(\dfrac{a^2-b^2}{a^2+b^2}\right)^2}\\\\\\=\pm\sqrt{1-\dfrac{(a^2-b^2)^2}{(a^2+b^2)^2}}\\\\\\=\pm\sqrt{\dfrac{(a^2+b^2)^2-(a^2-b^2)^2}{(a^2+b^2)^2}}\\\\\\=\pm\sqrt{\dfrac{4a^2b^2}{(a^2+b^2)^2}}\\\\\\=\pm\dfrac{2ab}{a^2+b^2}.$

So,

$\tan\theta=\dfrac{\sin\theta}{\cos \theta}=\dfrac{\dfrac{a^2-b^2}{a^2+b^2}}{\pm\dfrac{2ab}{a^2+b^2}}=\pm\dfrac{a^2-b^2}{2ab}.$

Again,

$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\dfrac{a^2-b^2}{a^2+b^2}}=\dfrac{a^2+b^2}{a^2-b^2},\\\\\\\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{1}{\pm\dfrac{2ab}{a^2+b^2}}=\pm\dfrac{a^2+b^2}{2ab},\\\\\\\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{\pm\dfrac{a^2-b^2}{2ab}}=\pm\dfrac{2ab}{a^2-b^2}.$

Thus, we found all the trigonometric ratios.