Question

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

Answer 1

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

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