Question

If sinA is a2 b2/ a2 b2



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Answer 1

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}.sinθ=

a2+b2 a2-b2

.

We are to find all the other trigonometric ratios.

We will be using the following relations among trigonometric ratios :

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

We have

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

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,

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

Thus, we found all the trigonometric ratios.