Question

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

• A. ${\cos }^{-1}\left\{\frac{14-\sqrt{5}}{3\sqrt{50}}\right\}$
• B. ${\cos }^{-1}\left\{\frac{10-\sqrt{5}}{3\sqrt{50}}\right\}$
• C. ${\cos }^{-1}\left\{\frac{14-\sqrt{15}}{3\sqrt{50}}\right\}$
• D. None of these

Answer 1

The correct option is A ${\cos }^{-1}\left\{\frac{14-\sqrt{5}}{3\sqrt{50}}\right\}$

We have,

${\cos }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{1}{7}$

We know that,

${\tan }^{-1}x={\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

Therefore,

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

$={\cos }^{-1}\frac{2}{3}+{\cos }^{-1}\left(\frac{7}{\sqrt{50}}\right)$

We know that,

${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\{xy-\sqrt{1-x^2}\sqrt{1-y^2}\}$

Therefore,

$={\cos }^{-1}\{\frac{2}{3}\times \frac{7}{\sqrt{50}}-\sqrt{1-{\left(\frac{2}{3}\right)}^2}\sqrt{1-{\left(\frac{7}{\sqrt{50}}\right)}^2}\}$

$={\cos }^{-1}\{\frac{14}{3\sqrt{50}}-\sqrt{\frac{5}{9}}\sqrt{\frac{1}{50}}\}$

$={\cos }^{-1}\{\frac{14}{3\sqrt{50}}-\frac{\sqrt{5}}{3\sqrt{50}}\}$

$={\cos }^{-1}\left\{\frac{14-\sqrt{5}}{3\sqrt{50}}\right\}$

Hence, this is the answer.