How to solve cos^-1(x^2-1/x^2+1)+tan^-1(2x/x^2-1)=2 /3 pi ?

Welcome to my article How to solve cos^-1(x^2-1/x^2+1)+tan^-1(2x/x^2-1)=2/3 pi ? 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 How to solve cos^-1(x^2-1/x^2+1)+tan^-1(2x/x^2-1)=2/3 pi ?, read and understand it carefully till the end.

Let us know how to solve this question How to solve cos^-1(x^2-1/x^2+1)+tan^-1(2x/x^2-1)=2/3 pi ?

First write the question on the page of the notebook,

cos^-1(x^2-1/x^2+1)+tan^-1(2x/x^2-1)=2/3 pi ?

Let us solve this question line by line in this way,

First of all write it like this, then let’s solve it.

\displaystyle {{\cos }^{{-1}}}\left( {\frac{{{{x}^{2}}-1}}{{{{x}^{2}}+1}}} \right)+{{\tan }^{{-1}}}\left( {\frac{{2x}}{{{{x}^{2}}-1}}} \right)=\frac{2}{3}\pi ,

\displaystyle {{\cos }^{{-1}}}\left( {-\frac{{\left( {1-{{x}^{2}}} \right)}}{{1+{{x}^{2}}}}} \right)+{{\tan }^{{-1}}}\left( {-\frac{{\left( {2x} \right)}}{{1-{{x}^{2}}}}} \right)=\frac{{2\pi }}{3} ,

since, FORMULA

\displaystyle {{\cos }^{{-1}}}\left( {-A} \right)=\pi -{{\cos }^{{-1}}}A , …._eq.(1) ,

AND

\displaystyle {{\tan }^{{-1}}}\left( {-A} \right)=-{{\tan }^{{-1}}}A , ….._eq.(2) ,

From eq. (1) and (2)

\displaystyle \pi -{{\cos }^{{-1}}}\left( {\frac{{1-{{x}^{2}}}}{{1+{{x}^{2}}}}} \right)-{{\tan }^{{-1}}}\left( {\frac{{2x}}{{1-{{x}^{2}}}}} \right)=\frac{{2\pi }}{3} ,

according to formula,

\displaystyle \pi -2{{\tan }^{{-1}}}x-2{{\tan }^{{-1}}}x=\frac{{2\pi }}{3} ,

\displaystyle \pi -\frac{{2\pi }}{3}=4{{\tan }^{{-1}}}x ,

\displaystyle \frac{{3\pi -2\pi }}{3}=4{{\tan }^{{-1}}}x ,

\displaystyle \frac{\pi }{3}=4{{\tan }^{{-1}}}x ,

\displaystyle \frac{\pi }{3}=\frac{{4{{{\tan }}^{{-1}}}x}}{1} ,

on slant multiplication ,

\displaystyle \frac{\pi }{{12}}={{\tan }^{{-1}}}x ,

or,

\displaystyle {{\tan }^{{-1}}}x=\frac{\pi }{{12}} ,

\displaystyle x=\tan \frac{\pi }{{12}} ,

\displaystyle x=\tan \left( {\frac{\pi }{4}-\frac{\pi }{6}} \right) ,

\displaystyle \because formula ,

\displaystyle \tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}} ,

\displaystyle x=\frac{{\tan \frac{\pi }{4}-\tan \frac{\pi }{6}}}{{1+\tan \frac{\pi }{4}\tan \frac{\pi }{6}}} ,

formula:-

\displaystyle \because \tan \frac{\pi }{4}=\tan {{45}^{\circ }}=1 ,

and ,

\displaystyle \because \tan \frac{\pi }{6}=\tan {{30}^{\circ }}=\frac{1}{{\sqrt{3}}} ,

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then,

\displaystyle x=\frac{{\tan {{{45}}^{\circ }}-\tan {{{30}}^{\circ }}}}{{1+\tan {{{45}}^{\circ }}\tan {{{30}}^{\circ }}}} ,

\displaystyle x=\frac{{1-\frac{1}{{\sqrt{3}}}}}{{1+1\times \frac{1}{{\sqrt{3}}}}} ,

\displaystyle x=\frac{{1-\frac{1}{{\sqrt{3}}}}}{{1+\frac{1}{{\sqrt{3}}}}} ,

\displaystyle x=\frac{{\frac{{\sqrt{3}-1}}{{\sqrt{3}}}}}{{\frac{{\sqrt{3}+1}}{{\sqrt{3}}}}} ,

\displaystyle x=\frac{{\sqrt{3}-1}}{{\sqrt{3}}}\times \frac{{\sqrt{3}}}{{\sqrt{3}+1}} ,

\displaystyle x=\frac{{\sqrt{3}-1}}{{\sqrt{3}+1}} ,

\displaystyle \text{Multiplying up and down by}(\sqrt{3}-1). ,

\displaystyle x=\frac{{\sqrt{3}-1}}{{\sqrt{3}+1}}\times \frac{{\sqrt{3}-1}}{{\sqrt{3}-1}} ,

\displaystyle x=\frac{{{{{\left( {\sqrt{3}-1} \right)}}^{2}}}}{{{{{\left( {\sqrt{3}} \right)}}^{2}}-{{{\left( 1 \right)}}^{2}}}} ,

we know that,

\displaystyle \left( {A-B} \right)\left( {A-B} \right)={{\left( {A-B} \right)}^{2}}

and,

\displaystyle \left( {A+B} \right)\left( {A-B} \right)={{\left( A \right)}^{2}}-{{\left( B \right)}^{2}} ,

then ,

\displaystyle x=\frac{{{{{\left( {\sqrt{3}} \right)}}^{2}}+{{{\left( 1 \right)}}^{2}}-2\times \sqrt{3}\times 1}}{{\left( {\sqrt{3}\times \sqrt{3}} \right)-\left( {1\times 1} \right)}} ,

\displaystyle x=\frac{{3+1-2\sqrt{3}}}{{3-1}} ,

\displaystyle x=\frac{{4-2\sqrt{3}}}{2} ,

\displaystyle x=\frac{{2\left( {2-\sqrt{3}} \right)}}{2} ,

\displaystyle x=\left( {2-\sqrt{3}} \right) ,

After solving line by line in this way, the solution of this question is-

\displaystyle x=\left( {2-\sqrt{3}} \right)\left[ {Answer} \right] ,

mathwaycalulus

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