Question

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

Answer 1

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 \Rightarrow5x + 3x = n\pi + \dfrac{\pi }{2}$$ $\Rightarrow8x = n\pi + \dfrac{\pi }{2} Now on taking L.CM for the above term where L.C.M = 2 we get $\Rightarrow$$$8x = \dfrac{{2n\pi + \pi }}{2}
Transposing 2 to left side and multiply with 8 we get
\Rightarrow$$$8x \times 2 = 2n\pi + \pi $$ \Rightarrow16x = 2n\pi + \pi $$ Transposing 16 on right side as denominator $\Rightarrowx = \dfrac{1}{{16}}(2\pi n + \pi )$Hence Required value of$x = \dfrac{1}{{16}}(2n\pi + \pi )$Or$x = \dfrac{\pi }{{16}}(2n + 1)$[Taking$\pi $$ 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).$$