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

Answer 1

Trigonometric Equations

We are given the trigonometric equation $\tan 5\alpha = \cot 3\alpha$ and we are supposed to solve it.

We will be using the following identities.

$\textsf{IDENTITIES}\\\boxed{\begin{minipage}{20em} \sf 2\ sin\, A\ sin\, B = cos\, (A-B) - \, cos\, (A+B) \\\\ 2\ cos\, A\ cos\, B = cos\, (A+B) + \, cos\, (A-B) \end{minipage}}$

Let's proceed to solving the equation:

$\displaystyle\tan 5\alpha=\cot 3\alpha\\\\\\ \implies\frac{\sin 5\alpha}{\cos 5\alpha}=\frac{\cos 3\alpha}{\sin 3\alpha}\\\\\\ \implies \sin 5\alpha\sin 3\alpha=\cos 5\alpha\cos 3\alpha \\\\\\ \textsf{Multipling 2 on both sides}\\\\\\ \implies 2\sin 5\alpha\sin 3\alpha = 2\cos 5\alpha\cos 3\alpha \\\\\\ \textsf{Use the identities} \\\\\\ \implies \cos (5\alpha-3\alpha)-\cos (5\alpha+3\alpha) = \cos(5\alpha+3\alpha)+\cos(5\alpha-3\alpha)$

$\implies \cancel{\cos 2\alpha} - \cos 8\alpha = \cos 8\alpha + \cancel{\cos 2\alpha} \\\\\\ \implies 2\cos 8\alpha =0 \\\\\\ \implies \cos 8\alpha = 0$

Now, when $\cos x = 0$, it means that x lies on the Y-axis, as an odd integral multiple of $\bf \frac{\pi}{2}$

$\cos x = 0 \implies x = n\pi+\dfrac{\pi}{2}\ ; n\in \mathbb{Z}$

Hence, here we have:

$\displaystyle\cos 8\alpha = 0\\\\\\ \implies 8\alpha = n\pi + \frac{\pi}{2}\ ; n\in \mathbb{Z} \\\\\\ \implies \boxed{\alpha = \frac{1}{8}\left(n\pi+\frac{\pi}{2}\right); n\in \mathbb{Z}}$

Here, we have $n\in \mathbb{N}$ and the answer is Option [A].

$\Large\boxed{\sf[A]\ \ \alpha = \frac{1}{8}\left(n\pi + \frac{\pi}{2}\right)}$

$\rule{300}{1}$

Shortcut Method

We can think smart by analyzing the options.

Options [B], [C] and [D] have some integral multiples of $\pi$. At such values of angles, tan becomes 0.

$\tan 5(n\pi) = 0 \\\\ \tan 5(2n\pi) = 0 \\\\ \tan 5(4n\pi)=0$

And cot is not even defined. So, Options [B], [C] and [D] cannot be correct. Hence, the Answer is Option [A].

Answer 2

Trigonometric Equations

We are given the trigonometric equation $\tan 5\alpha = \cot 3\alpha$ and we are supposed to solve it.

We will be using the following identities.

$\textsf{IDENTITIES}\\\boxed{\begin{minipage}{20em} \sf 2\ sin\, A\ sin\, B = cos\, (A-B) - \, cos\, (A+B) \\\\ 2\ cos\, A\ cos\, B = cos\, (A+B) + \, cos\, (A-B) \end{minipage}}$

Let's proceed to solving the equation:

$\displaystyle\tan 5\alpha=\cot 3\alpha\\\\\\ \implies\frac{\sin 5\alpha}{\cos 5\alpha}=\frac{\cos 3\alpha}{\sin 3\alpha}\\\\\\ \implies \sin 5\alpha\sin 3\alpha=\cos 5\alpha\cos 3\alpha \\\\\\ \textsf{Multipling 2 on both sides}\\\\\\ \implies 2\sin 5\alpha\sin 3\alpha = 2\cos 5\alpha\cos 3\alpha \\\\\\ \textsf{Use the identities} \\\\\\ \implies \cos (5\alpha-3\alpha)-\cos (5\alpha+3\alpha) = \cos(5\alpha+3\alpha)+\cos(5\alpha-3\alpha)$

$\implies \cancel{\cos 2\alpha} - \cos 8\alpha = \cos 8\alpha + \cancel{\cos 2\alpha} \\\\\\ \implies 2\cos 8\alpha =0 \\\\\\ \implies \cos 8\alpha = 0$

Now, when $\cos x = 0$, it means that x lies on the Y-axis, as an odd integral multiple of $\bf \frac{\pi}{2}$

$\cos x = 0 \implies x = n\pi+\dfrac{\pi}{2}\ ; n\in \mathbb{Z}$

Hence, here we have:

$\displaystyle\cos 8\alpha = 0\\\\\\ \implies 8\alpha = n\pi + \frac{\pi}{2}\ ; n\in \mathbb{Z} \\\\\\ \implies \boxed{\alpha = \frac{1}{8}\left(n\pi+\frac{\pi}{2}\right); n\in \mathbb{Z}}$

Here, we have $n\in \mathbb{N}$ and the answer is Option [A].

$\Large\boxed{\sf[A]\ \ \alpha = \frac{1}{8}\left(n\pi + \frac{\pi}{2}\right)}$

$\rule{300}{1}$

Shortcut Method

We can think smart by analyzing the options.

Options [B], [C] and [D] have some integral multiples of $\pi$. At such values of angles, tan becomes 0.

$\tan 5(n\pi) = 0 \\\\ \tan 5(2n\pi) = 0 \\\\ \tan 5(4n\pi)=0$

And cot is not even defined. So, Options [B], [C] and [D] cannot be correct. Hence, the Answer is Option [A].