Question

Prove that   cotA + cot(60+A) - cot(60-A) = 3cot3A .

Answer 1

hope this is help full

Answer 2

Answer and Explanation:

To prove : $\cot A+\cot (60+A)-\cot(60-A)=3\cot 3A$

Proof :

Taking LHS,

$LHS=\cot A+\cot (60+A)-\cot(60-A)$

Write, $\cot A=\frac{1}{\tan A}$

$LHS=\frac{1}{\tan A}+\frac{1}{\tan (60+A)}-\frac{1}{\tan (60-A)}$

Applying formula, $\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$

$LHS=\frac{1}{\tan A}+\frac{1-\tan 60\tan A}{\tan 60+\tan A}-\frac{1+\tan 60\tan A}{\tan 60-\tan A}$

$LHS=\frac{1}{\tan A}+\frac{1-\sqrt3\tan A}{\sqrt3+\tan A}-\frac{1+\sqrt3\tan A}{\sqrt3-\tan A}$

$LHS=\frac{1}{\tan A}+\frac{(1-\sqrt3\tan A)(\sqrt3-\tan A)-(1+\sqrt3\tan A)(\sqrt3+\tan A)}{(\sqrt3+\tan A)(\sqrt3-\tan A)}$

$LHS=\frac{1}{\tan A}+\frac{-8\tan A}{3-\tan^2 A}$

$LHS=\frac{3-\tan^2 A-8\tan^2 A}{(3-\tan^2 A)(\tan A)}$

$LHS=\frac{3(1-3\tan^2 A)}{(3-\tan^2 A)(\tan A)}$

$LHS=\frac{3}{\tan 3A}$

$LHS=3\cot 3A$

$LHS=RHS$

Hence Proved.