Question

Assertion :If $I_n=\int {\tan }^{n}xdx$, then $6(I_7+I_5)={\tan }^{6}x$ Reason: If $I_n=\int {\tan }^{n}xdx$ then $I_n=\frac{{\tan }^{n-1}x}{n}-I_{n-2}\forall n$

• 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 C Assertion is correct but Reason is incorrect

$I_n=\int {\tan }^{n-2}x({\sec }^{2}x-1)dx$
$=\int {\tan }^{n-2}x{\sec }^{2}xdx-I_{n-2}$
Substitute $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt$
$\therefore I_n=\int t^{n-2}dt-I_{n-2}$
$I_n=\frac{t^{n-1}}{n-1}-I_{n-2}$
$I_n=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}$ .....(1)
Substitute $x=7$, we have
$6(I_7+I_5)={\tan }^{6}x$ (*)
$\therefore$ Assertion (A) is true.
From (1), we can conclude reason is false.
Also, in Reason (R) if we substitute $n=7$, we get
$I_7+I_5=\frac{{\tan }^{6}x}{7}$
$\Rightarrow$ $7(I_7+I_5)={\tan }^{6}x$ which does not match with (*)
$\therefore$ Reason (R) is false.