Question

If $\sin x=\frac{a^2-b^2}{a^2+b^2}$, then the values of tan x, sec x and cosec x

Answer 1

$$\begin{gathered} \sin x=\frac{a^2-b^2}{a^2+b^2} \\ \mathrm{We}\mathrm{know}: \\ {\sin }^{2}x+{\cos }^{2}x=1 \\ {\cos }^{2}x=1-{\sin }^{2}x \\ =1-{\left(\frac{a^2-b^2}{a^2+b^2}\right)}^2 \\ =\frac{\left(a^4+b^4+2a^2{b}^2\right)-\left(a^4+b^4-2a^2{b}^2\right)}{{\left(a^2+b^2\right)}^2} \\ =\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow \cos x=\frac{2ab}{\left(a^2+b^2\right)} \end{gathered}$$

$$\begin{gathered} \tan x=\frac{\sin x}{\cos x}=\frac{{\displaystyle \frac{a^2-b^2}{a^2+b^2}}}{{\displaystyle \frac{2ab}{a^2+b^2}}}=\frac{a^2-b^2}{2ab} \\ \sec x=\frac{1}{\cos x}=\frac{a^2+b^2}{2ab} \\ \mathrm{cosec}x=\frac{1}{\sin x}=\frac{a^2+b^2}{a^2-b^2} \end{gathered}$$