Question

Pls help me out of this question

$\large\purple{Given}$

cotA -tanA = 2cot2A then

$\large\purple{To find}$

tanA - cotA +2tan2A + 4tan4A + 8cot8A

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Answer 1

EXPLANATION.

⇒ cot(A) - tan(A) = 2cot2A.

⇒ tan(A) - cot(A) + 2tan2A + 4tan4A + 8cot8A.

As we know that,

We can write equation as,

⇒ tan(A) - cot(A) = -2cot2A.  -----------(1).

Put the value of equation (1) in main equation, we get.

⇒ -2cot2A + 2tan2A + 4tan4A + 8cot8A.

As we know that,

⇒ tan2θ = 2tanθ/1 - tan²θ.

Using the formula in equation, we get.

We can write,

⇒ tanθ = 1/cotθ.

$\sf \implies \dfrac{-2}{tan(2A)} \ + 2tan2A \ + 4tan4A \ + \dfrac{8}{tan(8A)}$

$\sf \implies \dfrac{-2}{tan(2A)} \ + 2tan2A \ + 4tan4A \ + \dfrac{8}{tan \ 2(4A)}$

$\sf \implies \dfrac{-2(1 - tan^{2} A)}{2tanA} \ + 2tan2A \ + 4tan4A \ + \dfrac{8(1 - tan^{2}4A) }{2\ tan4A}$

$\sf \implies \dfrac{-(1 - tan^{2} A)}{tanA} \ + 2tan2A \ + 4tan4A \ + \dfrac{4(1 - tan^{2}4A) }{\ tan4A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + 2tan2A \ + \dfrac{4tan^{2}4A + 4(1 - tan^{2} 4A)}{tan4A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + 2tan2A \ + \dfrac{4tan^{2}4A + 4 - 4tan^{2} 4A}{tan4A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + 2tan2A \ + \dfrac{4}{tan4A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + 2tan2A + \dfrac{4}{tan \ 2(2A)}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + 2tan2A \ + \dfrac{4(1 - tan^{2} 2A)}{2 \ tan2A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + 2tan2A \ + \dfrac{2( 1 - tan^{2} 2A)}{tan2A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + \dfrac{2tan^{2}2A + 2(1 - tan^{2} 2A) }{tan2A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + \dfrac{2tan^{2}2A + 2 - 2tan^{2} 2A }{tan2A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + \dfrac{2}{tan2A}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + \dfrac{2(1 - tan^{2}A) }{2 tanA}$

$\sf \implies \dfrac{-(1 - tan^{2}A) }{tanA} \ + \dfrac{(1 - tan^{2}A) }{ tanA}$

$\sf \implies \dfrac{- 1 + tan^{2}A + 1 - tan^{2}A }{tanA} = 0.$

                                                                                                                         

MORE INFORMATION.

Some useful series.

(1) = sinα + sin(α + β) + sin(α + 2β) + ,,,,,to n terms.

$\sf \implies \dfrac{sin \bigg[\alpha + \bigg(\dfrac{n - 1}{2}\bigg)\beta \bigg] \bigg[sin \bigg(\dfrac{n\beta }{2} \bigg) \bigg] }{sin \bigg(\dfrac{\beta }{2}\bigg) }$

⇒ β ≠ 2nπ.

where,

⇒ α = initial angle.

⇒ β = difference angle.

⇒ n = no of terms.

(2) = cosα + cos(α + β) + cos(α + 2β) + ,,,,,,,to n terms.

$\sf \implies \dfrac{cos \bigg[\alpha + \bigg(\dfrac{n - 1}{2}\bigg)\beta \bigg] \bigg[sin \bigg(\dfrac{n\beta }{2} \bigg) \bigg] }{sin \bigg(\dfrac{\beta }{2}\bigg) }$

⇒ β ≠ 2nπ.