Question

Solve the following equation $cosec\theta =1+\cot \theta$.

Answer 1

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

Complete step-by-step answer:
We have been given the equation $cosec\theta =1+\cot \theta$.
As we know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{1}{\tan \theta }$.
So by substituting these values in the given equation, we get as follows:

& \dfrac{1}{\sin \theta }=1+\dfrac{1}{\tan \theta } \\\ & \dfrac{1}{\sin \theta }=1+\dfrac{\cos \theta }{\sin \theta } \\\ \end{aligned}$$ Now, taking the term $$\left( \dfrac{\cos \theta }{\sin \theta } \right)$$ to the left hand side of the equation, we get as follows: $$\dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta }=1$$ Taking $$\left( \dfrac{1}{\sin \theta } \right)$$ as common, we get as follows: $$\dfrac{1}{\sin \theta }\left( 1-\cos \theta \right)=1$$ On multiplying the equation by $$\sin \theta $$, we get as follows: $$\begin{aligned} & \dfrac{\sin \theta }{\sin \theta }\left( 1-\cos \theta \right)=\sin \theta \\\ & 1-\cos \theta =\sin \theta \\\ & \sin \theta +\cos \theta =1 \\\ \end{aligned}$$ Now we will multiply $$\dfrac{1}{\sqrt{2}}$$ to both the sides of the equation and then we will use the trigonometric identity which is as follows: $$\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$$ On multiplying the equation by $$\dfrac{1}{\sqrt{2}}$$, we get as follows: $$\dfrac{1}{\sqrt{2}}\sin \theta +\dfrac{1}{\sqrt{2}}\cos \theta =\dfrac{1}{\sqrt{2}}$$ Since we know that the value of $$\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$$, we get as follows: $$\begin{aligned} & \cos \dfrac{\pi }{4}\sin \theta +\sin \dfrac{\pi }{4}\cos \theta =\dfrac{1}{\sqrt{2}} \\\ & \sin \left( \theta +\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} \\\ & \sin \left( \theta +\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right) \\\ \end{aligned}$$ As we know that the general solution for $$\sin \theta =\sin \alpha $$ is given by $$\theta =n\pi +{{(-1)}^{n}}\alpha $$ where ‘n’ is any integer. $$\begin{aligned} & \sin \left( \theta +\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right) \\\ & \theta +\dfrac{\pi }{4}=n\pi +{{(-1)}^{n}}\dfrac{\pi }{4} \\\ & \theta =n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{4} \\\ & \theta =n\pi +\left[ {{(-1)}^{n}}-1 \right]\dfrac{\pi }{4} \\\ \end{aligned}$$ On substituting the obtained value of ‘$$\theta $$’ in the above equation we can observe that it doesn’t satisfy for $$n\in $$ even numbers. For $$n\in $$ even numbers, we get $$\infty =\infty $$. For $$n\in $$ odd numbers, we get 1 = 1. $$\begin{aligned} & \theta =n\pi 4\left( -1-1 \right)\dfrac{\pi }{4} \\\ & \theta =n\pi -\dfrac{\pi }{2} \\\ \end{aligned}$$ Therefore, the solution of the above equation is given by $$\theta =n\pi -\dfrac{\pi }{2}$$. Note: Don’t square the equation $$\sin \theta +\cos \theta =1$$ to solve the equation as it might give you the correct answer some times but it is not the correct method. Also, be careful when you get the solution for the equation. You must check it once by substituting it in the given equation.