If CosecA+cotA=m Show that - M …

Given:  cosec A + cot A = m
{tex}\Rightarrow{/tex} (cosec A + cot A)² = (m)²   [squaring both sides ]
{tex}\Rightarrow{/tex} cosec²A + cot²A + 2 cosec A cot A = m²  .......(1)
Now, LHS

 {tex}=\frac{m^{2}-1}{m^{2}+1}{/tex}
{tex}=\frac{\ cosec ^{2} A+\cot ^{2} A+2 \ cosec A \cot A-1}{\ cosec ^{2} A+\cot ^{2} A+2 \ cosce A \cdot \cot A+1}{/tex}. [ From (1) ]
{tex}=\frac{\cot ^{2} A+\cot ^{2} A+2 \ cosec A \cdot \cot A}{\ cosec ^{2} A+\ cosec ^{2} A+2 \ cosec A \cdot \cot A}{/tex}   [Since, Cosec²A - Cot²A = 1]
{tex}=\frac{2 \cot ^{2} A+2 \ cosec A \cot A}{2 \ cosec ^{2} A+2 \ cosec A \cot A}{/tex}
{tex}=\frac{2 \cot A(\cot A+\ cosec A)}{2 \ cosec A(\ cosec A+\cot A)}{/tex}
{tex}=\frac{\cot A}{cosec A}{/tex}
{tex}=\frac{\frac{\cos A}{\sin A}}{\frac{1}{\sin A}}{/tex}
{tex}=\frac{\cos A}{\sin A} \times \frac{\sin A}{1}{/tex}
= cos A = RHS

Hence, Proved.