SATHEE: Trigonometrical Ratios and Identities 1 Question 25

Trigonometrical Ratios and Identities 1 Question 25

25. Prove that

$\tan α + 2 \tan 2 α + 4 \tan 4 α + 8 \cot 8 α = \cot α (1988, 2 M)$

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Solution:

We know that,

$\cot θ - \tan θ = \frac{1 - \tan^{2} θ}{\tan θ} = 2 \frac{1 - \tan^{2} θ}{2 \tan θ} = 2 \cot 2 θ$

$L H S = \tan α + 2 \tan 2 α + 4 \tan 4 α + 8 \cot 8 α$

$\begin{array}{r} = - (\cot α - \tan α - 2 \tan 2 α - 4 \tan 4 α) \\ + 8 \cot 8 α + \cot α \\ = - (2 \cot 2 α - 2 \tan 2 α - 4 \tan 4 α) \\ + 8 \cot 8 α + \cot α \end{array}$

[from Eq. (i)]

$= - (2 (\cot 2 α - \tan 2 α) - 4 \tan 4 α)$ $+ 8 \cot 8 α + \cot α$

$= - (2 (2 \cot 4 α) - 4 \tan 4 α) + 8 \cot 8 α + \cot α$

[from Eq. (i)]

$= - 4 (\cot 4 α - \tan 4 α) + 8 \cot 8 α + \cot α$

$= - 8 \cot 8 α + 8 \cot 8 α + \cot α$

[from Eq. (i)]

$= \cot α = R H S$

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Trigonometrical Ratios and Identities 1 Question 25

25. Prove that

Solution:

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