Question

If $3\cot \theta =4$, show that $\frac{\left(1-{\tan }^2\theta \right)}{\left(1+{\tan }^2\theta \right)}=\left({\cos }^2\theta -{\sin }^2\theta \right)$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\left(1-{\tan }^2\theta \right)}{\left(1+{\tan }^2\theta \right)} \\ =\frac{\left(1-{\displaystyle \frac{1}{{\cot }^2\theta }}\right)}{\left(1+{\displaystyle \frac{1}{{\cot }^2\theta }}\right)} \\ =\frac{\left({\displaystyle \frac{{\cot }^2\theta -1}{{\cot }^2\theta }}\right)}{\left({\displaystyle \frac{{\cot }^2\theta +1}{{\cot }^2\theta }}\right)} \\ =\frac{{\cot }^2\theta -1}{{\cot }^2\theta +1} \\ =\frac{{\left({\displaystyle \frac{4}{3}}\right)}^2-1}{{\left({\displaystyle \frac{4}{3}}\right)}^2+1}\left(\mathrm{As},3\cot \theta =4\mathrm{or}\cot \theta =\frac{4}{3}\right) \end{gathered}$$
$$\begin{gathered} =\frac{{\displaystyle \frac{16}{9}-1}}{{\displaystyle \frac{16}{9}+1}} \\ =\frac{\left({\displaystyle \frac{16-9}{9}}\right)}{\left({\displaystyle \frac{16+9}{9}}\right)} \\ =\frac{\left({\displaystyle \frac{7}{9}}\right)}{\left({\displaystyle \frac{25}{9}}\right)} \\ =\frac{7}{25} \end{gathered}$$
$$\begin{gathered} \mathrm{RHS}=\left({\cos }^2\theta -{\sin }^2\theta \right) \\ =\frac{\left({\cos }^2\theta -{\sin }^2\theta \right)}{1} \\ =\frac{\left({\displaystyle \frac{{\cos }^2\theta -{\sin }^2\theta }{{\sin }^2\theta }}\right)}{\left({\displaystyle \frac{1}{{\sin }^2\theta }}\right)} \\ =\frac{\left({\displaystyle \frac{{\cos }^2\theta }{{\sin }^2\theta }}-{\displaystyle \frac{{\sin }^2\theta }{{\sin }^2\theta }}\right)}{{\mathrm{cosec}}^2\theta } \\ =\frac{\left({\cot }^2\theta -1\right)}{\left({\cot }^2\theta +1\right)} \end{gathered}$$
$$\begin{gathered} =\frac{\left[{\left({\displaystyle \frac{4}{3}}\right)}^2-1\right]}{\left[{\left({\displaystyle \frac{4}{3}}\right)}^2+1\right]} \\ =\frac{\left({\displaystyle \frac{16}{9}-\frac{1}{1}}\right)}{\left({\displaystyle \frac{16}{9}+\frac{1}{1}}\right)} \\ =\frac{\left({\displaystyle \frac{16-9}{9}}\right)}{\left({\displaystyle \frac{16+9}{9}}\right)} \\ =\frac{\left({\displaystyle \frac{7}{9}}\right)}{\left({\displaystyle \frac{25}{9}}\right)} \\ =\frac{7}{25} \end{gathered}$$

Since, LHS=RHS
Hence, verified.