Question

if A is acute angle , cotA+cosecA=3 then sinA=?​

Answer 1

$\textbf{Given:}$

$cotA+cosecA=3$

$\implies\displaystyle\frac{cosA}{sinA}+\frac{1}{sinA}=3$

$\implies\displaystyle\frac{1+cosA}{sinA}=3$

$\implies\displaystyle\frac{1+cosA}{sinA}=3$

$\implies\displaystyle\frac{1+2cos^2\frac{A}{2}-1}{2\;sin\frac{A}{2}\;cos\frac{A}{2}}=3$

$\implies\displaystyle\frac{2cos^2\frac{A}{2}}{2\;sin\frac{A}{2}\;cos\frac{A}{2}}=3$

$\implies\displaystyle\frac{cos\frac{A}{2}}{sin\frac{A}{2}}=3$

$\implies\displaystyle\;cot\frac{A}{2}=3$

$\text{Taking reciprocal on bothsides, we get}$

$\displaystyle\;tan\frac{A}{2}=\frac{1}{3}$

$\text{Now,}$

$sinA=\displaystyle\frac{2\;tan\frac{A}{2}}{1+tan^2\frac{A}{2}}$

$sinA=\displaystyle\frac{2(\frac{1}{3})}{1+\frac{1}{9}}$

$sinA=\displaystyle\frac{\frac{2}{3}}{\frac{10}{9}}$

$sinA=\displaystyle\frac{\frac{2}{3}}{\frac{10}{9}}$

$sinA=\displaystyle\frac{2}{3}{\times}\frac{9}{10}$

$sinA=\displaystyle\frac{1}{1}{\times}\frac{3}{5}$

$\therefore\bf\;sinA=\displaystyle\frac{3}{5}$

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