Question

Find the general solution for $\text{cosecx=1+cotx}$

Answer 1

Let us consider $\text{cosecx=1+cotx}$ as equation (1). We know that $\text{cosecx} =\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$. Now we have to substitute $\text{cosecx} =\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$ in equation (1). Now we have to divide the obtained equations into certain parts. Now for each part we have to calculate the solution separately. Now the union of all the solutions of different parts will give us a general solution of $\text{cosecx=1+cotx}$.

Complete step-by-step answer:
Let us assume
$\text{cosecx=1+cotx}.......\text{(1)}$
We know that $\text{cosecx}=\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$.
Now we have to substitute $\text{cosecx}=\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$ in equation (1).

& \Rightarrow \dfrac{1}{\operatorname{sinx}}=1+\dfrac{\cos x}{\sin x} \\\ & \Rightarrow \dfrac{1}{\sin x}=\dfrac{\sin x+\cos x}{\sin x} \\\ \end{aligned}$$ By cross multiplication, we get $$\begin{aligned} & \Rightarrow \sin x=\sin x(\sin x+\cos x) \\\ & \Rightarrow \sin x(1-(\sin x+\cos x))=0 \\\ & \Rightarrow \sin x=0\text{ (or) 1-(sinx+cosx)=0} \\\ \end{aligned}$$ Case 1: We know that $$\text{sinx=0}$$ if $$\text{x=n}\pi $$where n is equal to 0, 1, 2, 3, 4, 5,…….. $$\begin{aligned} & \Rightarrow \text{sinx=0} \\\ & \Rightarrow \text{x=n}\pi \\\ \end{aligned}$$ According to case 1, The general solution of $$\text{cosecx=1+cotx}$$ is equal to $$\text{x=n}\pi $$ where n is equal to 0, 1, 2, 3, 4, 5,……… Case 2: $$\begin{aligned} & \Rightarrow 1-(\sin x+\cos x)=0 \\\ & \Rightarrow \sin x+\cos x=1 \\\ & \Rightarrow \sin x=1-\cos x \\\ \end{aligned}$$ Now by squaring on both sides, we get $$\Rightarrow {{\sin }^{2}}x={{(1-\cos x)}^{2}}$$ We know that $$si{{n}^{2}}x+{{\cos }^{2}}x=1$$. $$\Rightarrow \left( 1-{{\cos }^{2}}x \right)={{(1-cosx)}^{2}}$$ We know that $${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$$. $$\begin{aligned} & \Rightarrow (1+\cos x)(1-\cos x)={{(1-\cos x)}^{2}} \\\ & \Rightarrow (1+\cos x)(1-\cos x)-{{(1-\cos x)}^{2}}=0 \\\ & \Rightarrow (1-\cos x)(1+\cos x-(1-\cos x))=0 \\\ \end{aligned}$$ Case 2(a): $$\begin{aligned} & \Rightarrow 1-\cos x=0 \\\ & \Rightarrow \cos x=1 \\\ \end{aligned}$$ We know that if $$\cos x=1$$ then $$x=2n\pi $$ where n is equal to 0, 1, 2, 3, 4, 5,…….. $$\Rightarrow x=2n\pi $$ According to case 2(a), The general solution of $$\text{cosecx=1+cotx}$$ is equal to $$x=2n\pi $$ where n is equal to 0, 1, 2, 3, 4, 5,……… Case 2(b): $$\begin{aligned} & \Rightarrow 1+\cos x-1+\cos x=0 \\\ & \Rightarrow 2\cos x=0 \\\ & \Rightarrow \operatorname{cosx}=0 \\\ \end{aligned}$$ We know that if $$\cos x=0$$ then $$x=(2n+1)\dfrac{\pi }{2}$$ where n is equal to 0, 1, 2, 3, 4, 5,…….. $$\Rightarrow x=(2n+1)\dfrac{\pi }{2}$$ According to case 2(b), The general solution of $$\text{cosecx=1+cotx}$$ is equal to $$x=(2n+1)\dfrac{\pi }{2}$$where n is equal to 0, 1, 2, 3, 4, 5,……… **So, from all the cases it is clear that the solution of $$\text{cosecx=1+cotx}$$ is $$x=2n\pi ,(2n+1)\dfrac{\pi }{2}$$.** **Note:** If we observe the step $$\sin x=\sin x(\sin x+\cos x)$$, here many students will cancel sin x on both sides and write directly $$\sin x+\cos x=1$$. Then the students may miss the solutions obtained from the equation $$\sin x=0$$. This will not give us a general solution in a proper manner. So, students should have a clear view of this concept of finding the general solutions. So, this type of misconception should be avoided.