Question

please answer the question..​

Answer 1

$LHS = cos^2x+cos^2(x+\frac{\pi }{3} ) + cos^2(x-\frac{\pi }{3})$

$= cos^2x + [cosx*cos\frac{\pi }{3} - sinx*sin\frac{\pi }{3}]^2+[cosx*cos\frac{\pi }{3}+sinx*sin\frac{\pi }{3}]^2$

$Let$

$cosx*cos\frac{\pi }{3} = \frac{cosx}{2}=A$

$sinx*sin\frac{\pi }{3} =\frac{\sqrt{3}sinx}{2}= B$

$LHS=cos^2x + [A-B]^2+[A+B]^2$

$= cos^2x + [A-B+A+B]^2-2[A-B][A+B]$

$= cos^2x+4A^2 - 2[A^2-B^2]$

$Substituting$ $for$ $A$ $and$ $B$

$= cos^2x + 4*\frac{cos^2x}{4}-2*[\frac{cos^2x}{4}-\frac{3sin^2x}{4}]$

$= 2cos^2x-\frac{cos^2x}{2}+\frac{3sin^2x}{2}$

$=\frac{3cos^2x}{2}+\frac{3sin^2x}{2}}$

$= \frac{3}{2}[cos^2x+sin^2x]$

$= \frac{3}{2} = RHS$

$Hence$ $Proved$

HOPE IT HELPS!!