Question

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

Answer 1

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. \[n = 0,\,\, \pm 1,\,\,\, \pm 2,\,\, \pm 3).\]