Question

Assertion :${\prod }_{r=1}^n(1+\sec 2^r\theta )=\tan 2^n\theta \cot \theta$ Reason: $\cos \theta \cos 2\theta \cos 2^2\theta ....\cos 2^{n-1}\theta =\frac{\sin \left(2^n\theta \right)}{2^n\sin \theta }$

• 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 false but Reason is true

Answer 1

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

${\prod }_{r=1}^n(1+\sec 2^r\theta )$
$={\prod }_{r=1}^n\left(\frac{1+\cos 2^r\theta }{\cos 2^r\theta }\right)$
$=\frac{{\prod }_{r=1}^{n}2{\cos }^2{2}^{r-1}\theta }{{\prod }_{r=1}^n\cos 2^r\theta }$ $(\because 1+\cos 2\theta =2{\cos }^2\theta )$
$=\frac{2^n({\prod }_{r=1}^n\cos 2^{r-1}\theta {)}^2}{\cos 2\theta \cos 2^2\theta \cos 2^3\theta ....\cos 2^n\theta }$
Multiplying and dividing by $\cos \theta$
$=\cos \theta \left[\frac{2^n(\cos \theta \cos 2\theta \cos 2^2\theta \cos 2^3\theta ....\cos 2^{n-1}\theta {)}^2}{\cos \theta \cos 2\theta \cos 2^2\theta \cos 2^3\theta ....\cos 2^{n-1}\theta \cos 2^n\theta }\right]$
$=\cos \theta \left[\frac{2^n(\cos \theta \cos 2\theta \cos 2^2\theta \cos 2^3\theta ....\cos 2^{n-1}\theta )}{\cos 2^n\theta }\right]$
$=\cos \theta \left[\frac{2^n(\frac{\sin \left(2^n\theta \right)}{2^n\sin \theta }}{\cos 2^n\theta }\right]$ ($\because \cos \theta \cos 2\theta \cos 2^2\theta ....\cos 2^{n-1}\theta =\frac{\sin 2^n\theta }{2^n\sin \theta }$)
$=\tan 2^n\theta \cot \theta$
$=$R.H.S of Assertion.
Hence, assertion is true.
Now consider,$\cos \theta \cos 2\theta \cos 2^2\theta ....\cos 2^{n-1}\theta$
Multiplying and dividing by $2\sin \theta$
$=\frac{1}{2\sin \theta }(2\sin \theta \cos \theta \cos 2\theta \cos 2^2\theta ....\cos 2^{n-1}\theta )$
$=\frac{1}{2\sin \theta }(\sin 2\theta \cos 2\theta \cos 2^2\theta ....\cos 2^{n-1}\theta )$
Now, multiply and divide by 2,
$=\frac{1}{2^2\sin \theta }(\sin 2^2\theta \cos 2^2\theta ....\cos 2^{n-1}\theta )$
Again multiplying and dividing by 2,
$=\frac{1}{2^3\sin \theta }(\sin 2^3\theta \cos 2^3\theta ....\cos 2^{n-1}\theta )$
Continuing this process, we get
$=\frac{1}{2^n\sin \theta }\left(\sin 2^{n-1}\theta \cos 2^{n-1}\theta \right)$
$=\frac{\sin \left(2^n\theta \right)}{2^n\sin \theta }$
Hence, reason is true and is the correct explanation for assertion.