Question

$\frac{3+\cot \left({76}^o\right).\cot \left(1^o\right)}{\cot \left({76}^o\right)+\cot \left({16}^o\right)}=\cot \left(\right(ab{)}^o)$ where $ab$ is a two digit natural number then $\frac{a\times b}{a+b}$ equals

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

$formulaused$

$1-\cos x=2{\sin }^2\frac{x}{2}$

$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$

$\cos (a-b)=\cos a\cos b+\sin a\sin b$

$\sin (a+b)=\sin a\cos b+\cos a\sin b$

$2\sin a\sin b=\cos (a-b)-\cos (a+b)$

$Given$

$\frac{3+\cot 76\cot 16}{\cot 76+\cot 16}=\cot ab$

$LHS$

$\frac{3+\cot 76\cot 16}{\cot 76+\cot 16}$

$=\frac{3+\frac{\cos 76}{\sin 76}\frac{\cos 16}{\sin 16}}{\frac{\cos 76}{\sin 76}+\frac{\cos 16}{\sin 16}}$

$=\frac{\frac{3\sin 76\sin 16+\cos 76\cos 16}{\sin 76\sin 16}}{\frac{\cos 76\sin 16+\cos 16\sin 76}{\sin 76\sin 16}}$

$=\frac{2\sin 76\sin 16+\cos s76\cos 16+\sin 76\sin 16}{\cos 76\sin 16+\cos 16\sin 76}$

$=\frac{2\sin 76\sin 16+\cos (76-16)}{\sin (76+16)}$

$=\frac{2\sin 76\sin 16+\cos 60}{\sin 92}$

$=\frac{\cos 60-\cos 92+\cos 60}{\sin 92}$

$=\frac{1-\cos 92}{\sin 92}$

$=\frac{2\sin 46}{2\sin 46\cos 46}$

$=\tan 46$

$=\cot (90-46)$

$=\cot 44$

$Hencea=4,b=4$

$\frac{a\times b}{a+b}$

$=\frac{4\left(4\right)}{4+4}$

$=\frac{16}{8}$

$=2$