Question

If $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$, find the values of all T-ratios of $\theta$.

Answer 1

We have $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$,

As,

$$\begin{gathered} {\cos }^2\theta =1-{\sin }^2\theta \\ =1-{\left(\frac{a^2-b^2}{a^2+b^2}\right)}^2 \\ =\frac{1}{1}-\frac{{\left(a^2-b^2\right)}^2}{{\left(a^2+b^2\right)}^2} \\ =\frac{{\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2}{{\left(a^2+b^2\right)}^2} \\ =\frac{\left[\left(a^2+b^2\right)-\left(a^2-b^2\right)\right]\left[\left(a^2+b^2\right)+\left(a^2-b^2\right)\right]}{{\left(a^2+b^2\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{\left[a^2+b^2-a^2+b^2\right]\left[a^2+b^2+a^2-b^2\right]}{{\left(a^2+b^2\right)}^2} \\ =\frac{\left[2b^2\right]\left[2a^2\right]}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow {\cos }^2\theta =\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow \cos \theta =\sqrt{\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2}} \\ \Rightarrow \cos \theta =\frac{2ab}{\left(a^2+b^2\right)} \end{gathered}$$

Also,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{\left({\displaystyle \frac{a^2-b^2}{a^2+b^2}}\right)}{\left({\displaystyle \frac{2ab}{a^2+b^2}}\right)} \\ =\frac{a^2-b^2}{2ab} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{cosec}\theta =\frac{1}{\sin \theta } \\ =\frac{1}{\left({\displaystyle \frac{a^2-b^2}{a^2+b^2}}\right)} \\ =\frac{a^2+b^2}{a^2-b^2} \end{gathered}$$

Also,

$$\begin{gathered} \sec \theta =\frac{1}{\cos \theta } \\ =\frac{1}{\left({\displaystyle \frac{2ab}{a^2+b^2}}\right)} \\ =\frac{a^2+b^2}{2ab} \end{gathered}$$

And,

$$\begin{gathered} \cot \theta =\frac{1}{\tan \theta } \\ =\frac{1}{\left({\displaystyle \frac{a^2-b^2}{2ab}}\right)} \\ =\frac{2ab}{a^2-b^2} \end{gathered}$$