Question

# If $\tan 5x = \cot 3x\,$then $x = (n \in z)$

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Hint: Trigonometric functions describe the relation between the sides and angles of a right triangle. ... The trigonometric functions include the following 6 functions: sine, cosine, tangent, cotangent, secant, and cosecant. For each of these functions, there is an inverse trigonometric function.
Always convert All the Trigonometric Function into the same function.
As to convert $\tan x$ into $\cot x$ the following ways are :
$\cot x = \tan (\dfrac{\pi }{2} - y)$
Formula for $\tan \theta = \tan \alpha$ this is the solution.
$\Rightarrow \theta = n\pi + \alpha$

Complete step-by- step solution:
Given $\tan 5x = \cot 3x$
Changing the $cot\theta$into $tan\theta$ by writing $cot\theta$= $tan\theta$
Where $\theta = 3x$
$\tan 5x = \tan (\dfrac{\pi }{2} - 3x)$
For General formula of $tan{\text{ }}x,$ when $tan\theta = tan{\text{ }}\alpha$
Then, $\theta = n\pi {\text{ }} + \alpha$
$\Rightarrow$$5x = n\pi + \dfrac{\pi }{2} - 3x$
Now on shifting $\pi /2$ on right side we get
$\Rightarrow$$5x + 3x = n\pi + \dfrac{\pi }{2}$
$\Rightarrow$$8x = n\pi + \dfrac{\pi }{2}$
Now on taking L.CM for the above term where L.C.M = 2 we get
$\Rightarrow$$8x = \dfrac{{2n\pi + \pi }}{2}$
Transposing 2 to left side and multiply with 8 we get
$\Rightarrow$$8x \times 2 = 2n\pi + \pi$
$\Rightarrow$$16x = 2n\pi + \pi$
Transposing 16 on right side as denominator
$\Rightarrow$$x = \dfrac{1}{{16}}(2\pi n + \pi )$
Hence Required value of $x = \dfrac{1}{{16}}(2n\pi + \pi )$
Or $x = \dfrac{\pi }{{16}}(2n + 1)$ [Taking $\pi$ common]

Note: Explanation for General solution of $\tan \theta$$= \tan \alpha$ is given by $\theta = n\pi + \alpha ,\,n \in z$.
As follows,
$\tan \theta = \tan \alpha$
We know $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow$ $\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\sin \alpha }}{{\cos \alpha }}$
Taking $\dfrac{{\sin \alpha }}{{\cos \alpha }}$ to L.H.S we get
$\Rightarrow$ $\dfrac{{\sin \theta }}{{\cos \theta }} - \dfrac{{\sin \alpha }}{{\cos \alpha }} = 0$
On taking taking LCM we get
$\Rightarrow$$\dfrac{{\sin \theta \cos \alpha - \sin \alpha \cos \theta }}{{\cos \theta \cos \alpha }} = 0$
$\Rightarrow$$\therefore \cos \theta \cos \alpha \times 0 = 0$
$\Rightarrow$$\sin \theta \cos \alpha - \sin \alpha \cos \theta = 0$
Identity: $\sin \theta \cos \alpha - \sin \alpha \cos \theta$ ,$\sin (\theta - \alpha )$
$\sin (\theta - \alpha ) = 0$
$\Rightarrow \theta - \alpha = n\pi$, where $n \in z$ i.e. $(n = 0, \pm 1, \pm 2, \pm 3.....)$.
(since we know that $\theta = n\pi ,n \in z$ is the general solution of the given equation $\sin \theta = 0$).
$\Rightarrow \theta = n\pi + \alpha ,$ where $n \in z$ (i.e. $n = 0,\,\, \pm 1,\,\,\, \pm 2,\,\, \pm 3).$