Question

Find the general solution of the equation of tanx=cotx

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{tan\,x=cot\,x}$

$\underline{\textbf{To find:}}$

$\textsf{The general solution of}\;\mathsf{tan\,x=cot\,x}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{tan\,x=cot\,x}$

$\mathsf{\dfrac{sin\,x}{cos\,x}=\dfrac{cos\,x}{sin\,x}}$

$\mathsf{sin^2x=cos^2x}$

$\mathsf{cos^2x-sin^2x=0}$

$\mathsf{cos\,2x=0}$

$\mathsf{cos\,2x=cos\,\dfrac{\pi}{2}}$

$\implies\mathsf{2x=2n\,\pi{\pm}\dfrac{\pi}{2}}$

$\implies\mathsf{x=n\,\pi{\pm}\dfrac{\pi}{4},\;\;\;n{\in}Z}$

$\textbf{The general solution of tanx=cotx is}$

$\bf\,x=n\,\pi{\pm}\dfrac{\pi}{4}$

$\underline{\textbf{Formula used:}}$

$\boxed{\begin{minipage}{5cm} \\\mathsf{Solution\;of\;cos\,\theta=cos\,\alpha\;is}\\\\\mathsf{\;\;\theta=2n\pi{\pm}\alpha,\;\;\;n{\in}Z}\\ \end{minipage}}$

Answer 2

Answer: x=$n\pi$±$\frac{\pi }{4}$

Step-by-step explanation:

• Given,

     tan x=cot x

  ⇒$\frac{sin x}{cos x} =\frac{cos x}{sin x}$

  ⇒$sin^{2}x=cos^{2}x$

  ⇒$cos^{2}x-sin^{2}x=0$

  ⇒ cos 2x=0

  ⇒ $cos 2x=cos \frac{\pi }{2}$

  ⇒$2x=\frac{\pi }{2}$

  ∴ x=$\frac{\pi }{4}$

So, x=$n\pi$±$\frac{\pi }{4}$ where n∈Z.

• Hence,the general solution for this given expression is x=$n\pi$±$\frac{\pi }{4}$.

     #SPJ2