Question

cot theta upon 1 + 10 square 90 minus theta minus sin theta into sin 90 minus theta​

Answer 1

$\textbf{Given:}$

$\displaystyle\frac{cot\,\theta}{1+tan^2(90-\theta)}-sin\,\theta\,sin(90-\theta)$

$\text{Using,}\bf\;tan(90-\theta)=cot\,\theta\;\text{and}\;sin(90-\theta)=cos\theta$

$=\displaystyle\frac{cot\,\theta}{1+cot^2\theta}-sin\,\theta\,cos\theta$

$\text{Using,}\bf\;cosec^2A=1+cot^2A$

$=\displaystyle\frac{cot\,\theta}{cosec^2\theta}-sin\,\theta\,cos\theta$

$=\displaystyle\frac{cos\,\theta}{sin\,\theta}(sin^2\theta)-sin\,\theta\,cos\theta$

$=\displaystyle\,sin\,\theta\,cos\,\theta-sin\,\theta\,cos\theta$

$=0$

$\therefore\displaystyle\frac{cot\,\theta}{1+tan^2(90-\theta)}-sin\,\theta\,sin(90-\theta)=0$