Question

Prove : $\cot {16}^o\cot {44}^o+\cot {44}^o\cot {76}^o-\cot {76}^o\cot {16}^o=3$.

Answer 1

$\cot {16}^{\circ }\cot {44}^{\circ }+\cot {44}^{\circ }\cot {76}^{\circ }+\cot {16}^{\circ }\cot {76}^{\circ }\cot (90-74)\cot (90-46)+\cot (90-46)\cot (90-14)+\cot (90-14)\cot (90-74)\tan {74}^{\circ }\tan {46}^{\circ }+\tan {46}^{\circ }\tan {14}^{\circ }+\tan {14}^{\circ }\tan {74}^{\circ }$

now

$\tan A+B=\frac{\tan A+\tan B}{1-\tan A\tan B}\tan A\tan B=1-\left[\frac{\tan A+\tan B}{\tan (A+B)}\right]\tan {74}^{\circ }\tan {46}^{\circ }=1-\left[\frac{\tan {74}^{\circ }+\tan {46}^{\circ }}{\tan (74+46{)}^{\circ }}\right]=1-\left[\frac{\tan {74}^{\circ }+\tan {46}^{\circ }}{\tan (120{)}^{\circ }}\right]=1-\left[\frac{\tan {74}^{\circ }+\tan {46}^{\circ }}{-\cot {30}^{\circ }}\right]=1-\left[\frac{\tan {74}^{\circ }+\tan {46}^{\circ }}{-\sqrt{3}}\right]=1+\frac{\tan {74}^{\circ }}{-\sqrt{3}}+\frac{\tan {46}^{\circ }}{-\sqrt{3}}$

similarly

$\tan {16}^{\circ }\tan {14}^{\circ }=1+\frac{\tan {16}^{\circ }}{-\sqrt{3}}+\frac{\tan {14}^{\circ }}{-\sqrt{3}}$

again

$\tan {74}^{\circ }\tan {14}^{\circ }=-1+\frac{\tan {74}^{\circ }}{\sqrt{3}}+\frac{\tan {14}^{\circ }}{-\sqrt{3}}$

by adding all above equations we get

$\tan {74}^{\circ }\tan {46}^{\circ }+\tan {46}^{\circ }\tan {14}^{\circ }+\tan {14}^{\circ }\tan {74}^{\circ }$

$=1+\frac{\tan {74}^{\circ }}{-\sqrt{3}}+\frac{\tan {46}^{\circ }}{-\sqrt{3}}+1+\frac{\tan {16}^{\circ }}{-\sqrt{3}}+\frac{\tan {14}^{\circ }}{-\sqrt{3}}-1+\frac{\tan {74}^{\circ }}{\sqrt{3}}+\frac{\tan {14}^{\circ }}{-\sqrt{3}}$

$=3$

LHS=RHS