If (cosec A + cot A) = m then prove that

Given: $\operatorname{cosec} A+\cot A=m$  ....(1)

We know

$\operatorname{cosec}^{2} A-\cot ^{2} A=1$

$\Rightarrow(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)=1 \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$

$\Rightarrow(\operatorname{cosec} A-\cot A) m=1 \quad[$ From $(1)]$

$\Rightarrow \operatorname{cosec} A-\cot A=\frac{1}{m} \quad \ldots \ldots(2)$

Adding (1) and (2), we get

$\operatorname{cosec} A+\cot A+\operatorname{cosec} A-\cot A=m+\frac{1}{m}$

$\Rightarrow 2 \operatorname{cosec} A=\frac{m^{2}+1}{m}$

$\Rightarrow \operatorname{cosec} A=\frac{m^{2}+1}{2 m}$

$\Rightarrow \frac{1}{\operatorname{cosec} A}=\frac{2 m}{m^{2}+1}$

$\Rightarrow \sin A=\frac{2 m}{m^{2}+1} \quad \ldots .(3)$

Subtracting (2) from (1), we get

$\operatorname{cosec} A+\cot A-\operatorname{cosec} A+\cot A=m-\frac{1}{m}$

$\Rightarrow 2 \cot A=\frac{m^{2}-1}{m}$

$\Rightarrow \cot A=\frac{m^{2}-1}{2 m} \quad \ldots \ldots(4)$

Now,

$\cos A=\sin A \times \cot A$

$\Rightarrow \cos A=\frac{2 m}{m^{2}+1} \times \frac{m^{2}-1}{2 m} \quad[$ From $(3)$ and $(4)]$

$\Rightarrow \cos A=\frac{m^{2}-1}{m^{2}+1}$