Question

If (cosec A + cot A) = m then prove that $\frac{m^2-1}{m^2+1}=\cos \mathrm{\theta }$.

Answer 1

Given: $\mathrm{cosec}A+\cot A=m.....\left(1\right)$
We know
$$\begin{gathered} {\mathrm{cosec}}^{2}A-{\cot }^{2}A=1 \\ \Rightarrow \left(\mathrm{cosec}A-\cot A\right)\left(\mathrm{cosec}A+\cot A\right)=1\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \\ \Rightarrow \left(\mathrm{cosec}A-\cot A\right)m=1\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow \mathrm{cosec}A-\cot A=\frac{1}{m}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} \mathrm{cosec}A+\cot A+\mathrm{cosec}A-\cot A=m+\frac{1}{m} \\ \Rightarrow 2\mathrm{cosec}A=\frac{m^2+1}{m} \\ \Rightarrow \mathrm{cosec}A=\frac{m^2+1}{2m} \\ \Rightarrow \frac{1}{\mathrm{cosec}A}=\frac{2m}{m^2+1} \end{gathered}$$
$\Rightarrow \sin A=\frac{2m}{m^2+1}.....\left(3\right)$
Subtracting (2) from (1), we get
$$\begin{gathered} \mathrm{cosec}A+\cot A-\mathrm{cosec}A+\cot A=m-\frac{1}{m} \\ \Rightarrow 2\cot A=\frac{m^2-1}{m} \\ \Rightarrow \cot A=\frac{m^2-1}{2m}.....\left(4\right) \end{gathered}$$
Now,
$$\begin{gathered} \cos A=\sin A\times \cot A \\ \Rightarrow \cos A=\frac{2m}{m^2+1}\times \frac{m^2-1}{2m}\left[\mathrm{From}\left(3\right)\mathrm{and}\left(4\right)\right] \\ \Rightarrow \cos A=\frac{m^2-1}{m^2+1} \end{gathered}$$