Question

tan inverse (1/root3 tanx/2)= 1/2 cos inverse((1+2cos x) /2+cos x)

Answer 1

tan^-1{(1/√3)tan(x/2)}

=(1/2) 2tan^-1{(1/√3)tan(x/2)}

=(1/2)cos^-1[{(1-(1/3)tan²(x/2)}/{(1+(1/3)tan²(x/2)}]

=(1/2)cos^-1[{3-tan²(x/2)}/{3+tan²(x/2)}]

=(1/2)cos^-1[{3-(1-cosx)/(1+cosx)}/{3+(1-cosx)/(1+cosx)}]

=(1/2)cos^-1{(2+4cosx)/(4+2cosx)}

=(1/2)cos^-1{(1+2cosx)/(2+cosx)}(Proved)

Answer 2

Answer:

Equation is verified.

Left Hand side  = Right hand side

Step-by-step explanation:

GIVEN:

$tan^{-1}(\frac{1}{\sqrt{3}}* tan\frac{x}{2} } ) = \frac{1}{2} cos^{-1}(\frac{(1+2cosx)}{(2+cosx)})$

We have to verify the equation.

from right hand side, we will transform the equation as

We know that

$cos^{-1}x = 2tan^{-1}(\frac{\sqrt{1-x^{2}} }{1+x})$

rewriting right hand side as

$\frac{1}{2} * 2 tan^{-1}(\frac{\sqrt{1 - \frac{(1+2cosx)^{2}}{(2+cosx)^{2}} } }{1 + \frac{1 + 2cosx}{2+cosx} } )$

then equation becomes

$tan^{-1}(\frac{\sqrt{4 + cos^2x + 4cosx - 1 - 4cos^2x - 4cosx} }{2 + cosx + 1 + 2cosx} )\\tan^{-1}(\frac{\sqrt{3 - 3cos^2x} }{3 + 3cosx } )\\\\tan^{-1}(\frac{\sqrt{3(1 - cos^2x)} }{3(1 + cosx) } )\\\\\\tan^{-1}(\frac{\sqrt{3sin^2x} }{3(1 + cosx)} )\\tan^{-1}(\frac{sinx\sqrt{3} }{3(1 + cosx)} )$

On comparing with left hand side we get $tan^{-1}$ eliminated

and equation becomes

$\frac{1}{\sqrt{3} } * tan\frac{x}{2} = \sqrt{3}sinx * \frac{1}{3(1+cosx)} \\ tan\frac{x}{2} = 3 sinx * \frac{1}{3(1+cosx)}\\ tan\frac{x}{2} = sinx * \frac{1}{1+cosx}$

rewriting equation as

$tan\frac{x}{2} = 2*sin\frac{x}{2} * cos\frac{x}{2} * \frac{1}{2 cos^2\frac{x}{2} } \\tan\frac{x}{2} = sin\frac{x}{2} * \frac{1}{cos\frac{x}{2} } \\ tan\frac{x}{2} = tan\frac{x}{2}$

LEFT HAND SIDE = RIGHT HAND SIDE

Hence, equation is verified