Proved that 2 tan^(-1)sqrt((b)/(a))=cos^(-1)(a-b)/(a+b) ?

Welcome to my article2 tan^(-1)sqrt((b)/(a))=cos^(-1)(a-b)/(a+b) ? This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question 2 tan^(-1)sqrt((b)/(a))=cos^(-1)(a-b)/(a+b) ?, read and understand it carefully till the end.

Let us know how to solve this question 2 tan^(-1)sqrt((b)/(a))=cos^(-1)(a-b)/(a+b) ?.

First write the question on the page of the notebook.

2 tan^(-1)sqrt((b)/(a))=cos^(-1)(a-b)/(a+b) ?

let’s solve this question,

LEFT HAND SIDE

\displaystyle 2ta{{n}^{{(-1)}}}sqrt((b)/(a))

On writing it like this –

\displaystyle 2ta{{n}^{{(-1)}}}\sqrt{{\left( {\frac{{\left( b \right)}}{{\left( a \right)}}} \right)}}

if,

\displaystyle 2ta{{n}^{{(-1)}}}\sqrt{{\left( {\frac{{\left( b \right)}}{{\left( a \right)}}} \right)}}=\theta

\displaystyle 2\sqrt{{\left( {\frac{{\left( b \right)}}{{\left( a \right)}}} \right)}}=tan\theta

\displaystyle 2\left( {\frac{{\left( b \right)}}{{\left( a \right)}}} \right)=ta{{n}^{2}}\theta

so that we write

\displaystyle 2\theta

\displaystyle {{\cos }^{2}}\theta =\frac{{1-ta{{n}^{2}}\theta }}{{1+ta{{n}^{2}}\theta }}

\displaystyle 2\theta ={{\cos }^{{-1}}}\left( {\frac{{1-ta{{n}^{2}}\theta }}{{1+ta{{n}^{2}}\theta }}} \right)

\displaystyle 2\theta ={{\cos }^{{-1}}}\left( {\frac{{1-\frac{b}{a}}}{{1+\frac{b}{a}}}} \right)

\displaystyle 2\theta ={{\cos }^{{-1}}}\left( {\frac{{\frac{{a-b}}{a}}}{{\frac{{a+b}}{a}}}} \right)

\displaystyle 2\theta ={{\cos }^{{-1}}}\left( {\frac{{a-b}}{{a+b}}} \right)

\displaystyle ={{\cos }^{{-1}}}\left( {\frac{{a-b}}{{a+b}}} \right) (proved)

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