Question

How do you solve $\sec x+\csc x=0$?

Answer 1

The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. Trigonometric identities are the equations involving the trigonometric functions that are true for every value of the variables involved.

Complete step by step answer:
As per the given question, we need to solve the given trigonometric expression using trigonometric identities and algebraic formulae. Here, we are given the expression $\sec x+\csc x=0$.
We know that secant is the reciprocal of cosine function. And cosecant is the reciprocal of sine function. Now we can rewrite the equation as
$\dfrac{1}{\cos x}+\dfrac{1}{\sin x}=0$
Now we take the LCM of denominators then the equation will be
$\Rightarrow \dfrac{\sin x+\cos x}{\cos x\sin x}=0$
Now we multiply with $\cos x\sin x$ on both sides. Then we get

& \Rightarrow \dfrac{\sin x+\cos x}{\cos x\sin x}\times \cos x\sin x=0\times \cos x\sin x \\\ & \Rightarrow \sin x+\cos x=0 \\\ \end{aligned}$$ Now we multiply with $$\dfrac{1}{\sqrt{2}}$$ on both sides of the equation. then we get $$\begin{aligned} & \Rightarrow \dfrac{1}{\sqrt{2}}\left( \sin x+\cos x \right)=0\times \dfrac{1}{\sqrt{2}} \\\ & \Rightarrow \dfrac{1}{\sqrt{2}}\left( \sin x+\cos x \right)=0 \\\ \end{aligned}$$ $$\Rightarrow \dfrac{1}{\sqrt{2}}\sin x+\dfrac{1}{\sqrt{2}}\cos x=0$$ We know that according to trigonometric values, $$\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$$. So, on substituting these values, we can rewrite the equation as $$\Rightarrow \cos \dfrac{\pi }{4}\sin x+\sin \dfrac{\pi }{4}\cos x=0$$ We know that according to properties of trigonometric functions $$\sin (x+y)=\sin x\cos y+\cos x\sin y$$. So we can write the above equation as $$\Rightarrow \sin \left( x+\dfrac{\pi }{4} \right)=0$$. The value of sine function will be 0 only when $$\left( x+\dfrac{\pi }{4} \right)=0$$ or $$\left( x+\dfrac{\pi }{4} \right)=\pi $$. $$\begin{aligned} & \Rightarrow \left( x+\dfrac{\pi }{4} \right)=0\to x=\dfrac{-\pi }{4} \\\ & \Rightarrow \left( x+\dfrac{\pi }{4} \right)=\pi \to x=\pi -\dfrac{\pi }{4}=\dfrac{3\pi }{4} \\\ \end{aligned}$$ **Therefore, $$x=\dfrac{-\pi }{4},\dfrac{3\pi }{4}$$ solves the given equation $$\sec x+\csc x=0$$.** **Note:** In order to solve such types of questions, we need to have enough knowledge over trigonometric functions and identities. We also need to know the algebraic formulae to simplify the expressions. While making solutions of trigonometric value check all possible cases in order to get all the solutions. We must avoid calculation mistakes to get the correct solution.