Question

If tan(x-y) /2,tan z, tan(x+y) /2 are in G. P. prove that cos x=cosy*cos2z​

Answer 1

Answer

• cos x = cos y × cos 2z

Given

• tan(x-y)/2 , tan z , tan(x+y)/2 are in G. P

To prove

• cos x=cosy * cos2z​

Proof

If a , b , c are in GP

then , b² = ac

Here , tan(x-y)/2,tan z, tan(x+y)/2 are in G. P

$\implies \bf tan^2z=\bigg(tan\dfrac{x-y}{2}\bigg).\bigg(tan\dfrac{x+y}{2}\bigg)\\\\\implies \bf \dfrac{sin^2z}{cos^2z}=\dfrac{sin\dfrac{x-y}{2}.sin\dfrac{x+y}{2}}{cos\dfrac{x-y}{2}.cos\dfrac{x+y}{2}}\\\\\bf applying\ componendo\ and\ dividendo\ method,we\ get ,\\\\\implies \bf \dfrac{sin^2x+cos^2x}{sin^2x-cos^2x}=\dfrac{sin\dfrac{x-y}{2}.sin\dfrac{x+y}{2}+cos\dfrac{x-y}{2}.cos\dfrac{x+y}{2}}{sin\dfrac{x-y}{2}.sin\dfrac{x+y}{2}-cos\dfrac{x-y}{2}.cos\dfrac{x+y}{2}}$

$\implies \bf \dfrac{1}{-cos2x}=\dfrac{cos(\dfrac{x-y}{2}-\dfrac{x+y}{2})}{-cos(\dfrac{x-y}{2}+\dfrac{x+y}{2})}\\\\\implies \bf -cosx=cos(-y)*-cos2x\\\\\implies \bf cosx=cosy*cos2z\;[\;Since\ cos(-x)=cosx\;]$

Hence proved