Question

If n is a natural number, which of the following is the solution to the equation tan(5α)=cot(3α)?

[A] α=1/8 (nπ+π/2) [B] α=nπ [C] α=2nπ [D] α=4nπ

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Answer 1

Step-by-step explanation:

Trigonometric Equations

We are given the trigonometric equation and we are supposed to solve it.

We will be using the following identities.

Let's proceed to solving the equation:

Now, when , it means that x lies on the Y-axis, as an odd integral multiple of

Hence, here we have:

Here, we have and the answer is Option [A].

Answer 2

$\large\underline{\sf{Solution-}}$

Given Trigonometric equation is

$\rm :\longmapsto\:tan5 \alpha = cot3 \alpha$

can be rewritten as

$\rm :\longmapsto\:\dfrac{sin5 \alpha }{cos5 \alpha } = \dfrac{cos3 \alpha }{sin3 \alpha }$

$\rm :\longmapsto\:cos5 \alpha \: cos3 \alpha = sin5 \alpha \: sin3 \alpha$

$\rm :\longmapsto\:cos5 \alpha \: cos3 \alpha - sin5 \alpha \: sin3 \alpha = 0$

We know,

$\boxed{\tt{ cosx \: cosy \: - \: sinx \: siny \: = \: cos(x + y) \: }}$

So, using this identity, we get

$\rm :\longmapsto\:cos(5 \alpha + 3 \alpha ) = 0$

$\rm\implies \:cos8 \alpha = 0$

We know,

$\boxed{\tt{ cosx = 0 \: \rm\implies \:x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z}}$

So, using this identity,

$\rm\implies \:8 \alpha = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: N$

can be rewritten as

$\rm\implies \:8 \alpha = n\pi + \dfrac{\pi}{2}\: \forall \: n \in \: N$

$\rm\implies \:\alpha =\dfrac{1}{8} \bigg( n\pi + \dfrac{\pi}{2}\bigg)\: \forall \: n \in \: N$

So, Option [ A ] is correct.

ALTERNATIVE METHOD

Given Trigonometric equation is

$\rm :\longmapsto\:tan5 \alpha = cot3 \alpha$

can be rewritten as

$\rm :\longmapsto\:tan5 \alpha = tan\bigg(\dfrac{\pi}{2} - 3 \alpha\bigg)$

$\rm\implies \:5 \alpha =n \pi + \dfrac{\pi}{2} - 3 \alpha \: \: \: \forall \: n \in \: N$

$\rm\implies \:8 \alpha = n\pi + \dfrac{\pi}{2}\: \forall \: n \in \: N$

$\rm\implies \:\alpha =\dfrac{1}{8} \bigg( n\pi + \dfrac{\pi}{2}\bigg)\: \forall \: n \in \: N$

So, option [ A ] is correct.

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Additional information

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$