Question

Assertion :If $\alpha ,\beta$ are the roots of the equation $18(ta{n}^{-1}x{)}^2-9\pi ta{n}^{-1}x+{\pi }^2=0$ then $\alpha +\beta =\frac{4}{\sqrt{3}}$. Reason: $se{c}^2\left(co{s}^{-1}\left(\frac{1}{4}\right)\right)+cose{c}^2\left(si{n}^{-1}\left(\frac{1}{5}\right)\right)=41$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Reason:
${\sec }^2\left({\cos }^{-1}\frac{1}{4}\right)+{\csc }^2\left(si{n}^{-1}\frac{1}{5}\right)={\sec }^2\left({\sec }^{-1}4\right)+{\csc }^2\left({\csc }^{-1}5\right)=16+25=41$
Assertion:
$18{\left({\tan }^{-1}x\right)}^2-9\pi {\tan }^{-1}x+{\pi }^2=0$

$\Rightarrow {\left({\tan }^{-1}x\right)}^2-\frac{\pi }{2}{\tan }^{-1}x+\frac{{\pi }^2}{18}=0$

$\Rightarrow {\left({\tan }^{-1}x\right)}^2-(\frac{\pi }{6}+\frac{\pi }{3}){\tan }^{-1}x+(\frac{\pi }{6}\times \frac{\pi }{3})=0$

$\left({\tan }^{-1}\right(x)-\frac{\pi }{6})\left({\tan }^{-1}\right(x)-\frac{\pi }{3})=0$

Given $\alpha ,\beta$ are the roots$\Rightarrow {\tan }^{-1}\alpha =\frac{\pi }{6},{\tan }^{-1}\beta =\frac{\pi }{3}$

$\alpha =\tan \frac{\pi }{6}=\sqrt{3},\beta =\tan \frac{\pi }{3}=\frac{1}{\sqrt{3}}$

$\therefore \alpha +\beta =\tan \frac{\pi }{6}+\tan \frac{\pi }{3}=\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{4}{\sqrt{3}}$

Hence,both Assertion and Reason are correct but Reason is not the correct explanation for Assertion