Question

Solve the following equation \[\cot \theta +\tan \theta =2\].

Answer 1

Hint: As we know that tangent of any angle is the ratio of sine is to cosine of the angle and cotangent is reciprocal of tangent. So we will substitute the values of cotangent and tangent in terms of sine and cosine in the given equation.

Complete Step-by-step answer:
We have been given the equation \[\cot \theta +\tan \theta =2\].
As we know \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\].
So by substituting these values of \[\tan \theta \] and \[\cot \theta \] in the given equation, we get as follows:
\[\dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta }=2\]
Now we can take the LCM of the terms, and then we will get as follows:
\[\begin{align}
  & \dfrac{\cos \theta \times \cos \theta +\sin \theta \times \sin \theta }{\sin \theta .\cos \theta }=2 \\
 & \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\sin \theta .\cos \theta }=2 \\
\end{align}\]
As we know that \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]
So, by substituting the value of \[\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\] in the above equation, we get as follows:
\[\begin{align}
  & \dfrac{1}{\sin \theta .\cos \theta }=2 \\
 & 1=2\sin \theta .\cos \theta \\
 & \therefore 2\sin \theta .\cos \theta =1 \\
\end{align}\]
Also, we know the trigonometric identity \[2\sin \theta .\cos \theta =\sin 2\theta \].
So, by using the identity \[\left( 2\sin \theta .\cos \theta \right)\] in the above equation, we get as follows:
\[\sin 2\theta =1\]
As we know that the general solution for \[\sin \theta =\sin \alpha \] is given by \[\theta =2n\pi +\alpha \], where n = 0, 1, 2, 3…
\[\begin{align}
  & \sin 2\theta =\sin \dfrac{\pi }{2} \\
 & 2\theta =2n\pi +\dfrac{\pi }{2} \\
 & \theta =\dfrac{1}{2}\left( 2n\pi +\dfrac{\pi }{2} \right) \\
 & \theta =n\pi +\dfrac{\pi }{4} \\
\end{align}\]
Therefore, the solution for the given equation is \[\theta =\left( n\pi +\dfrac{\pi }{4} \right)\].

Note: We can also solve the given equation by substituting \[\cot \theta \] to \[\dfrac{1}{\tan \theta }\] and then solve the quadratic equation in \[\tan \theta \].
But using this method can be time consuming as we have to find the roots and then the value of theta. So it would be better to convert to sine and cosine terms and find the value of theta.
Also, we must know the general solution of \[\sin \theta =\sin \alpha \] as it is really helpful in solving these types of questions.