Question

Find the value of $sin \bigg( 2 \ tan^{-1} \frac{2}{3} \bigg)+ cos\bigg(tan^{-1} \sqrt{3}\bigg)$

Answer 1

$\underline{\text{Solution :}}$

$\mathrm{Now,\:2tan^{-1}\frac{2}{3}}$

$\mathrm{=tan^{-1}\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^{2}}}$

$\mathrm{=tan^{-1}\frac{\frac{4}{3}}{1-\frac{4}{9}}}$

$\mathrm{=tan^{-1}\frac{\frac{4}{3}}{\frac{9-4}{9}}}$

$\mathrm{=tan^{-1}(\frac{4}{3}*\frac{9}{5})}$

$\mathrm{=tan^{-1}\frac{12}{5}}$

$\mathrm{=sin^{-1}\frac{12}{\sqrt{12^{2}+5^{2}}}}$

$\mathrm{=sin^{-1}\frac{12}{\sqrt{169}}}$

$\mathrm{=sin^{-1}\frac{12}{13}}$

$\to \mathrm{2tan^{-1}\frac{2}{3}=sin^{-1}\frac{12}{13}}$

$\to \mathrm{sin(2tan^{-1}\frac{2}{3})=sin(sin^{-1}\frac{12}{13})}$

$\to \mathrm{sin(2tan^{-1}\frac{2}{3})=\frac{12}{13}}$

$\mathrm{Then,\:sin(2tan^{-1}\frac{2}{3})+cos(tan^{-1}\sqrt{3})}$

$\mathrm{=\frac{12}{13}+cos\frac{\pi}{3}}$

$\mathrm{=\frac{12}{13}+\frac{1}{2}}$

$\mathrm{=\frac{24+13}{26}}$

$\mathrm{=\frac{37}{26}}$

$\to \boxed{\mathrm{sin(2tan^{-1}\frac{2}{3})+cos(tan^{-1}\sqrt{3})=\frac{37}{26}}}$

$\underline{\text{Rules :}}$

$\mathrm{1.\:2tan^{-1}x=tan^{-1}\frac{2x}{1-x^{2}}}$

$\mathrm{2.\:tan^{-1}\frac{x}{y}=sin^{-1}\frac{x}{\sqrt{x^{2}+y^{2}}}}$

Answer 2

ANSWER:--------

{2tan−132 }

{=tan−12(23)1−(23)2\mathrm}

{=tan^{-1}\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^{2}}}

{=tan−11−(32)22(32)} 

{=tan−1431−49\mathrm}

{=tan^{-1}\frac{\frac{4}{3}}{1-\frac{4}{9}}}{=tan−11−9434}

=tan−1439−49\mathrm{=tan^{-1}\frac{\frac{4}{3}}{\frac{9-4}{9}}}=tan−199−434 

=tan−1(43∗95)\mathrm{=tan^{-1}(\frac{4}{3}*\frac{9}{5})}=tan−1(34∗59) 

=tan−1125\mathrm{=tan^{-1}\frac{12}{5}}=tan−1512 

=sin−112122+52\mathrm{=sin^{-1}\frac{12}{\sqrt{12^{2}+5^{2}}}}=sin−1122+5212 

=sin−112169\mathrm{=sin^{-1}\frac{12}{\sqrt{169}}}=sin−116912 

=sin−11213\mathrm{=sin^{-1}\frac{12}{13}}=sin−11312 

→2tan−123=sin−11213\to \mathrm{2tan^{-1}\frac{2}{3}=sin^{-1}\frac{12}{13}}→2tan−132=sin−11312 

→sin(2tan−123)=sin(sin−11213)\to \mathrm{sin(2tan^{-1}\frac{2}{3})=sin(sin^{-1}\frac{12}{13})}→sin(2tan−132)=sin(sin−11312) 

→sin(2tan−123)=1213\to \mathrm{sin(2tan^{-1}\frac{2}{3})=\frac{12}{13}}→sin(2tan−132)=1312 

Then,sin(2tan−123)+cos(tan−13)\mathrm{Then,\:sin(2tan^{-1}\frac{2}{3})+cos(tan^{-1}\sqrt{3})}Then,sin(2tan−132)+cos(tan−13)

=1213+cosπ3\mathrm{=\frac{12}{13}+cos\frac{\pi}{3}}=1312+cos3π 

=1213+12\mathrm{=\frac{12}{13}+\frac{1}{2}}=1312+21 

=24+1326\mathrm{=\frac{24+13}{26}}=2624+13 

=3726\mathrm{=\frac{37}{26}}=2637 

→sin(2tan−123)+cos(tan−13)=3726\to \boxed{\mathrm{sin(2tan^{-1}\frac{2}{3})+cos(tan^{-1}\sqrt{3})=\frac{37}{26}}}→sin(2tan−132)+cos(tan−13)=2637

Rules :‾\underline{\text{Rules :}}Rules : 

1.2tan−1x=tan−12x1−x2\mathrm{1.\:2tan^{-1}x=tan^{-1}\frac{2x}{1-x^{2}}}1.2tan−1x=tan−11−x22x 

2.tan−1xy=sin−1xx2+y2\mathrm{2.\:tan^{-1}\frac{x}{y}=sin^{-1}\frac{x}{\sqrt{x^{2}+y^{2}}}}2.tan−1yx=sin−1x2+y2

HOPE IT HELPS:--------

T!—!ANKS!!!!