Question

How do i prove that Tan.Tan(π/3+A).Tan(π/3-A)=Tan3A?

Answer 1

Concept

Trigonometric equations can be solved to determine the angles by substituting basic formulas where possible.

Given

A trigonometric equation TanA Tan(π/3+A).Tan(π/3-A)=Tan3A?

Find

Whether LHS = RHS or not.

Solution

we know,   $\frac{\pi }{3}$ = 60°

Thus LHS becomes $Tan\frac{\pi }{3} . Tan{(\frac{\pi }{3}-x)}. Tan{(\frac{\pi }{3}+x)}$

=Tan(60°) . Tan(60°-x) . Tan(60°+x)

= $\frac{sin60.sin(60-x)sin(60+x)}{cos60.cos(60-x).cos(60+x)}$

=$\frac{sin x (sin^{2} 60 - sin^{2}x )}{cos x(cosx^{2} -sinx^{2} x)}$

=$\frac{3sinx - 4sinx^{3 }}{4cosx^{3} - 3cosx}$

=sin3x/cos3x

=tan3x

=RHS

LHS = RHS , Hence proved

#SPJ3