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Question: How do you prove \(\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x}=\sin x\csc x\)?...

How do you prove sin⁡xcsc⁡x+cos⁡xsec⁡x=sin⁡xcsc⁡x\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x}=\sin x\csc x?

Explanation

Solution

In this problem we need to prove that the given equation is true. For this we need to show that value of LHS=LHS=value of RHSRHS. So, we will consider both LHS and RHS individually. Now we will use the basic trigonometric formulas that are sin⁡x=1csc⁡x\sin x=\dfrac{1}{\csc x}, cos⁡x=1sec⁡x\cos x=\dfrac{1}{\sec x}. We will substitute those values in the LHS and simplify the equation by using the trigonometric formulas. After all simplifications we will get a value for the LHS. Now we will move to RHS and here also we will use the basic trigonometric formula sin⁡x=1csc⁡x\sin x=\dfrac{1}{\csc x} and simplify the equation to get the value of RHS. After getting the values of LHS and RHS we will compare them to show that the given equation is true or false.

Complete step-by-step solution:
Given equation, sin⁡xcsc⁡x+cos⁡xsec⁡x=sin⁡xcsc⁡x\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x}=\sin x\csc x.
The left-hand side part of the given equation is sin⁡xcsc⁡x+cos⁡xsec⁡x\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x} and the right-hand side part of the given equation is sin⁡xcsc⁡x\sin x\csc x.
Considering the Left-hand side part
LHS=sin⁡xcsc⁡x+cos⁡xsec⁡xLHS=\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x}
We have the basic trigonometric formulas sin⁡x=1csc⁡x\sin x=\dfrac{1}{\csc x}, cos⁡x=1sec⁡x\cos x=\dfrac{1}{\sec x}. Substituting these values in the LHS part, then we will get
LHS=sin⁡x(sin⁡x)+cos⁡x(cos⁡x)LHS=\sin x\left( \sin x \right)+\cos x\left( \cos x \right)
We know that a×a=a2a\times a={{a}^{2}}, then the LHS is modified as
LHS=sin⁡2x+cos⁡2xLHS={{\sin }^{2}}x+{{\cos }^{2}}x
We have the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, then the value of LHS will be
∴LHS=1\therefore LHS=1
Now considering the Right-hand side part of the given equation which is
RHS=sin⁡xcsc⁡xRHS=\sin x\csc x
We have a basic trigonometric formula sin⁡x=1csc⁡x\sin x=\dfrac{1}{\csc x}. From this formula the value of RHS will be
RHS=csc⁡xcsc⁡x ⇒RHS=1 \begin{aligned} & RHS=\dfrac{\csc x}{\csc x} \\\ & \Rightarrow RHS=1 \\\ \end{aligned}
We have LHS=RHS=1LHS=RHS=1. From this we can write sin⁡xcsc⁡x+cos⁡xsec⁡x=sin⁡xcsc⁡x\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x}=\sin x\csc x. Hence proved.

Note: For this problem we have followed a simple method by using the basic trigonometric formula. There are so many methods which differ in simplification only. For example, we can multiply the term sin⁡xcsc⁡x\dfrac{\sin x}{\csc x} with sin⁡xsin⁡x\dfrac{\sin x}{\sin x} and the term cos⁡xsec⁡x\dfrac{\cos x}{\sec x} with cos⁡xcos⁡x\dfrac{\cos x}{\cos x}. Now we will substitute the formula sin⁡xcsc⁡x=1\sin x\csc x=1, cos⁡xsec⁡x=1\cos x\sec x=1 and use the same trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 to get the value of LHS. For the RHS we will follow the same method.