The correct option is B 2α 1 tan 3α #160; tanα #160; #160; 1 1 tan 3α #160; #160; 1 tanα #160; #160; #160; 1 tan 3α ...

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Trigonometric Identities

Solve: 1tan3...

Question

Solve:
$\frac{1}{tan 3 α - tan α} - \frac{1}{cot 3 α - cot α} =$

0

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cos 2 α

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tan 2 α

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cot 2 α

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Solution

The correct option is B $cot 2 α$
$\frac{1}{tan 3 α - tan α} - \frac{1}{\frac{1}{tan 3 α} - \frac{1}{tan α}}$
$\frac{1}{tan 3 α - tan α} - \frac{tan 3 α tan α}{tan α - tan 3 α}$
$\frac{1}{tan 3 α - tan α} + \frac{tan 3 α tan α}{tan 3 α - tan α} = \frac{1}{tan 2 α} = cot 2 α$

Suggest Corrections

Similar questions

tan α + cot α = p

{tan}^{2} α + {cot}^{2} α

\tan α = \frac{1 - \cos β}{\sin β}

\tan 3 α = \tan 2 β

\tan 2 α = \tan β

\tan 2 β = \tan α

a = {sec}^{2} α, b = {c o s e c}^{2} α, a n d, c = \frac{1}{1 - {sin}^{2} α {cos}^{2} α} .

\frac{tan 3 α}{tan α} = \frac{λ a}{b}, t h e n λ =

α

∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + α)^{2} & (1 + 2 α)^{2} & (1 + 3 α)^{2} (2 + α)^{2} & (2 + 2 α)^{2} & (2 + 3 α)^{2} (3 + α)^{2} & (3 + 2 α)^{2} & (3 + 3 α)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 648 α

α^{t}

(α + p)^{m - 1} + (α + p)^{m - 2} (α + q) + (α + p)^{m - 3} (α + q)^{2} + . . . . . . . . (α + q)^{m - 1},

α \neq - q

p \neq q

(α + p)^{m - 1} + (α + p)^{m - 2} (α + q) + (α + p)^{m - 3} (α + q)^{2} + . . . . . . . . (α + q)^{m - 1},

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