Trigonometrical Ratios and Identities 1 Question 21
21. For a positive integer $n$, let
$$ \begin{aligned} & f _n(\theta)=\tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2 \theta) \\ & \left(1+\sec 2^{2} \theta\right) \ldots \quad\left(1+\sec 2^{n} \theta\right), \text { then } \end{aligned} $$
(1999, 3M)
(a) $f _2 \frac{\pi}{16}=1$
(b) $f _3 \frac{\pi}{32}=1$
(c) $f _4 \frac{\pi}{64}=1$
(d) $f _5 \frac{\pi}{128}=1$
Match the Column
Match the conditions/expressions in Column I with values in Column II.
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Solution:
- NOTE Multiplicative loop is very important approach in IIT Mathematics.
$$ \begin{aligned} & \begin{aligned} \tan \frac{\theta}{2}(1+\sec \theta) & =\frac{\sin \theta / 2}{\cos \theta / 2} \cdot 1+\frac{1}{\cos \theta} \\ & =\frac{(\sin \theta / 2) 2 \cos ^{2} \theta / 2}{(\cos \theta / 2) \cos \theta} \\ & =\frac{(2 \sin \theta / 2) \cos \theta / 2}{\cos \theta}=\frac{\sin \theta}{\cos \theta}=\tan \theta \\ \therefore \quad f _n(\theta) & =(\tan \theta / 2)(1+\sec \theta) \\ (1+ & \sec 2 \theta)\left(1+\sec 2^{2} \theta\right) \ldots\left(1+\sec 2^{n} \theta\right) \\ & =(\tan \theta)(1+\sec 2 \theta)\left(1+\sec 2^{2} \theta\right) \ldots\left(1+\sec 2^{n} \theta\right) \\ & =\tan 2 \theta \cdot\left(1+\sec 2^{2} \theta\right) \ldots\left(1+\sec 2^{n} \theta\right) \\ & =\tan \left(2^{n} \theta\right) \end{aligned} \end{aligned} $$
Now, $\quad f _2 \frac{\pi}{16}=\tan 2^{2} \cdot \frac{\pi}{16}=\tan \frac{\pi}{4}=1$
Therefore, (a) is the answer.
$$ f _3 \frac{\pi}{32}=\tan 2^{3} \cdot \frac{\pi}{32}=\tan \frac{\pi}{4}=1 $$
Therefore, (b) is the answer.
$$ f _4 \frac{\pi}{64}=\tan 2^{4} \cdot \frac{\pi}{64}=\tan \frac{\pi}{4}=1 $$
Therefore, (c) is the answer.
$$ f _5 \frac{\pi}{128}=\tan 2^{5} \cdot \frac{\pi}{128}=\tan \frac{\pi}{4}=1 $$
Therefore, (d) is the answer.
