Trigonometrical Ratios and Identities 3 Question 3
4. If $\tan A=(1-\cos B) / \sin B$, then $\tan 2 A=\tan B$.
(1993, 1M)
Analytical & Descriptive Questions
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Solution:
- Let $y=\frac{\sin x \cos 3 x}{\sin 3 x \cos x}=\frac{\tan x}{\tan 3 x}$
$$ \begin{array}{r} \Rightarrow \quad y=\frac{\tan x}{\tan 3 x}=\frac{\tan x\left(1-3 \tan ^{2} x\right)}{3 \tan x-\tan ^{3} x} \\ =\frac{1-3 \tan ^{2} x}{3-\tan ^{2} x} \end{array} $$
Put $\tan x=t$
$$ \begin{array}{ll} \Rightarrow & y=\frac{1-3 t^{2}}{3-t^{2}} \\ \Rightarrow & 3 y-t^{2} y=1-3 t^{2} \\ \Rightarrow & 3 y-1=t^{2} y-3 t^{2} \\ \Rightarrow & 3 y-1=t^{2}(y-3) \\ \Rightarrow & \frac{3 y-1}{y-3}=t^{2} \Rightarrow \frac{3 y-1}{y-3}>0 \\ \therefore & t^{2}>0 \Rightarrow \end{array} $$
NOTE It is a brilliant technique to convert equation into inequation and asked in IIT papers frequently. $\quad \Rightarrow y<1 / 3$ or $y>3$. This shows that $y$ cannot lie between $1 / 3$ and 3 .
