Trigonometrical Ratios and Identities 1 Question 25
25. Prove that
$\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha=\cot \alpha \quad(1988,2 M)$
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Solution:
- We know that,
$$ \cot \theta-\tan \theta=\frac{1-\tan ^{2} \theta}{\tan \theta}=2 \frac{1-\tan ^{2} \theta}{2 \tan \theta}=2 \cot 2 \theta $$
$$ LHS=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha $$
$$ \begin{array}{r} =-(\cot \alpha-\tan \alpha-2 \tan 2 \alpha-4 \tan 4 \alpha) \\ +8 \cot 8 \alpha+\cot \alpha \\ =-(2 \cot 2 \alpha-2 \tan 2 \alpha-4 \tan 4 \alpha) \\ +8 \cot 8 \alpha+\cot \alpha \end{array} $$
[from Eq. (i)]
$=-(2(\cot 2 \alpha-\tan 2 \alpha)-4 \tan 4 \alpha)$ $+8 \cot 8 \alpha+\cot \alpha$
$=-(2(2 \cot 4 \alpha)-4 \tan 4 \alpha)+8 \cot 8 \alpha+\cot \alpha$
[from Eq. (i)]
$=-4(\cot 4 \alpha-\tan 4 \alpha)+8 \cot 8 \alpha+\cot \alpha$
$=-8 \cot 8 \alpha+8 \cot 8 \alpha+\cot \alpha$
[from Eq. (i)]
$$ =\cot \alpha=RHS $$
