Question

Assertion :${\int }_{\pi /6}^{\pi /3}\frac{1dx}{1+\sqrt{\cot x}}=\frac{\pi }{2}$ Reason: ${\int }_{\pi /6}^{\pi /3}\frac{f\left(x\right)dx}{f\left(x\right)+f(\frac{\pi }{2}-x)}=\frac{\pi }{12}$

• 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. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{f\left(x\right)dx}{f\left(x\right)+f(\frac{\pi }{2}-x)}$ (i)
$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{f(\frac{\pi }{6}+\frac{\pi }{3}-x)}{f(\frac{\pi }{6}+\frac{\pi }{3}-x)+f(\frac{\pi }{2}-\frac{\pi }{6}+\frac{\pi }{3}-x)}dx$
$\therefore$ $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{f(\frac{\pi }{2}-x)}{f(\frac{\pi }{2}-x)+f\left(x\right)}$ (ii)
Adding (i) and (ii), we get
$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}1.dx=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$ $\therefore$ $I=\frac{\pi }{12}$
$\therefore$ Reason (R) is true
Again, for $f\left(x\right)=\frac{1}{1+\sqrt{\cot x}}=\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}$
$\therefore$ ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}f\left(x\right)dx=\frac{\pi }{12}=I$
$\therefore$ Assertion (A) is also true by use of Reason (R).