2{tan ^{ - 1}}left( {frac{1}{3}} right) + {ta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We use the formula $2{	an ^{ - 1}}(x) = {	an ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} ight)$ for $|x| < 1$. First,simplify $2{	an ^{ - 1}}\l

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(D) We use the formula $2{	an ^{ - 1}}(x) = {	an ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} ight)$ for $|x| First,simplify $2{	an ^{ - 1}}\left( {\frac{1}{3}} ight)$: $2{	an ^{ - 1}}\left( {\frac{1}{3}} ight) = {	an ^{ - 1}}\left( {\frac{{2(1/3)}}{{1 - {{(1/3)}^2}}}} ight) = {	an ^{ - 1}}\left( {\frac{{2/3}}{{1 - 1/9}}} ight) = {	an ^{ - 1}}\left( {\frac{{2/3}}{{8/9}}} ight) = {	an ^{ - 1}}\left( {\frac{2}{3} 	imes \frac{9}{8}} ight) = {	an ^{ - 1}}\left( {\frac{3}{4}} ight)$. Now,add the remaining term using the formula ${	an ^{ - 1}}(x) + {	an ^{ - 1}}(y) = {	an ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} ight)$: ${	an ^{ - 1}}\left( {\frac{3}{4}} ight) + {	an ^{ - 1}}\left( {\frac{1}{7}} ight) = {	an ^{ - 1}}\left( {\frac{{3/4 + 1/7}}{{1 - (3/4)(1/7)}}} ight)$. Calculate the numerator and denominator: Numerator: $\frac{3}{4} + \frac{1}{7} = \frac{{21 + 4}}{{28}} = \frac{{25}}{{28}}$. Denominator: $1 - \frac{3}{{28}} = \frac{{28 - 3}}{{28}} = \frac{{25}}{{28}}$. Thus,${	an ^{ - 1}}\left( {\frac{{25/28}}{{25/28}}} ight) = {	an ^{ - 1}}(1) = \frac{\pi }{4}$.

$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = $

Similar Questions

If $\cot^{-1} \alpha + \cot^{-1} \beta = \cot^{-1} x$,then $x = $

The value of $\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x$ is

$\tan \left[ {\frac{\pi }{4} + \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] + \tan \left[ {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] = $

If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$

If $0 < x < \frac{1}{\sqrt{2}}$ and $\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}$,then a value of $\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)$ is $......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module