Question

Assertion :If $x=\sqrt{{\cos }^2\theta +\sqrt{{\cos }^2\theta {\sqrt{\cos }}^2\theta +...\infty }}$ and $y=\sqrt{\sec \theta -\sqrt{\sec \theta -\sqrt{\sec \theta -...\infty }}}$ then $(x^2-x)(y^2+y)=1$ Reason: If $z=\sqrt{\tan \theta +\sqrt{\tan \theta +\sqrt{\tan \theta +...\infty }}}$ then $z^2=\tan \theta +z$.

• 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 D Assertion is incorrect but Reason is correct

Reason:

$z=\sqrt{\tan \theta +\sqrt{\tan \theta +\sqrt{\tan \theta +...\infty }}}\Rightarrow z=\sqrt{\tan \theta +z}\Rightarrow z^2=\tan \theta +z\Rightarrow z^2-z=\tan \theta$

Assertion:

Using reason we get

$x=\sqrt{{\cos }^2\theta +\sqrt{{\cos }^2\theta +\sqrt{{\cos }^2\theta +...\infty }}}$

$\Rightarrow x^2-x={\cos }^2\theta$ ...(1)

$y=\sqrt{sec\theta -\sqrt{\sec \theta -\sqrt{\sec \theta -...\infty }}}$

$\Rightarrow y^2+y=\sec \theta$ ...(2)

Therefore, from (1) and (2), we get

$(x^2-x)(y^2+y)={\cos }^2\theta \sec \theta =\cos \theta$