IIT-JEE 1999Question 1
For a positive integer $n$, let $f_n(\theta)=\left(\tan\dfrac{\theta}{2}\right)(1+\sec\theta)(1+\sec2\theta)(1+\sec4\theta)\cdots(1+\sec2^{n}\theta)$. Then
Trigonometric Ratio & Identities
Solution
Concept. Each factor $1+\sec2x$ turns $\tan x$ into $\tan2x$, so the product collapses to $\tan(2^n\theta)$. Formulas used. (1) $\tan x\,(1+\sec2x)=\tan2x$Step 1 - Why (1). $1+\sec2x=\dfrac{1+\cos2x}{\cos2x}=\dfrac{2\cos^2x}{\cos2x}$, so $$\tan x\,(1+\sec2x)=\frac{2\sin x\cos x}{\cos2x}=\tan2x.$$ Step 2 - Collapse. By (1) with $x=\frac\theta2,\theta,2\theta,\dots,2^{n-1}\theta$: $$f_n(\theta)=\tan\left(2^n\theta\right).$$ Step 3 - Check each option. $2^2\cdot\frac{\pi}{16}=2^3\cdot\frac{\pi}{32}=2^4\cdot\frac{\pi}{64}=2^5\cdot\frac{\pi}{128}=\frac\pi4$, and $\tan\frac\pi4=1$. All four are true. Answer: (A), (B), (C), (D)