Question

Qs. Prove that 2tan-1 [ { (a-b) / (a+b) }1/2 tan x/2 ] = cos-1{ (b +acos x ) / (a+bcos x) }

Answer 1

$\textbf{To prove:}$

$\bf\displaystyle2\,tan^{-1}[\sqrt{\frac{a-b}{a+b}}\frac{1}{2}tan\frac{x}{2}]=cos^{-1}[\frac{b+a\,cosx}{a+b\,cosx}]$

$\text{Take,}$

$A=cos^{-1}[\displaystyle\frac{b+a\,cosx}{a+b\,cosx}]$ .........(1)

$\implies\,cosA=\displaystyle\frac{b+a\,cosx}{a+b\,cosx}$

$\text{we know that,}$

$\boxed{\bf\;cosA=\frac{1-tan^2\frac{A}{2}}{1+tan^2\frac{A}{2}}\implies\;tan^2\frac{A}{2}=\frac{1-cosA}{1+cosA}}$

$\displaystyle\,tan^2\frac{A}{2}=\frac{1-(\frac{b+a\,cosx}{a+b\,cosx})}{1+(\frac{b+a\,cosx}{a+b\,cosx})}$

$\displaystyle\,tan^2\frac{A}{2}=\frac{a+b\,cosx-b-a\,cosx}{a+b\,cosx+b+a\,cosx}$

$\displaystyle\,tan^2\frac{A}{2}=\frac{(a-b)-(a-b)cosx}{(a+b)+(a+b)cosx}$

$\displaystyle\,tan^2\frac{A}{2}=\frac{(a-b)[1-cosx]}{(a+b)[1+cosx]}$

$\displaystyle\,tan^2\frac{A}{2}=(\frac{a-b}{a+b})(\frac{1-cosx}{1+cosx})$

$\displaystyle\,tan^2\frac{A}{2}=(\frac{a-b}{a+b})tan^2\frac{x}{2}$

$\text{Taking square root on both sides, we get}$

$\displaystyle\,tan\frac{A}{2}=\sqrt{\frac{a-b}{a+b}}tan\frac{x}{2}$

$\implies\displaystyle\frac{A}{2}=tan^{-1}[\sqrt{\frac{a-b}{a+b}}tan\frac{x}{2}]$

$\implies\displaystyle\,A=2\,tan^{-1}[\sqrt{\frac{a-b}{a+b}}tan\frac{x}{2}]$........(2)

$\text{From (1) and (2), we get}$

$\boxed{\bf\displaystyle2\,tan^{-1}[\sqrt{\frac{a-b}{a+b}}\frac{1}{2}tan\frac{x}{2}]=cos^{-1}[\frac{b+a\,cosx}{a+b\,cosx}]}$

Find more:

1.2cot inverse(7)+cos inverse (3/5)

https://brainly.in/question/3447900

2.Prove that tan inverse 1/2+tan inverse 1/5=1/2 cos inverse 16/65

https://brainly.in/question/14008410