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

How do you provesec⁡xcot⁡x+tan⁡x=sin⁡x\dfrac{{\sec x}}{{\cot x + \tan x}} = \sin x?

Explanation

Solution

The very first step to solve this problem, we will replace all the terms on the LHS with the required simple terms. After doing this, we will have sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1on the LHS numerator of the denominator. We will replace this with 1 because we know thatsin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1. We will solve and simplify the LHS in such a way that all the terms in the LHS will get cancelled out and we will end up with the value on the RHS. Therefore proving our question.

Formula Used: Use the given below formulas to solve the problem

cot⁡θ=1tan⁡θ sin⁡θ=1csc⁡θ cos⁡θ=1sec⁡θ  \cot \theta = \dfrac{1}{{\tan \theta }} \\\ \sin \theta = \dfrac{1}{{\csc \theta }} \\\ \cos \theta = \dfrac{1}{{\sec \theta }} \\\

Complete step-by-step solution:
The given question we have issec⁡xcot⁡x+tan⁡x=sin⁡x\dfrac{{\sec x}}{{\cot x + \tan x}} = \sin x
Before heading on to solve the above problem. We need to keep in mind few basic trigonometric formulas which will help you in almost every problem that you will face in this chapter.
And those special formulas are:-

cot⁡θ=1tan⁡θ sin⁡θ=1csc⁡θ cos⁡θ=1sec⁡θ  \cot \theta = \dfrac{1}{{\tan \theta }} \\\ \sin \theta = \dfrac{1}{{\csc \theta }} \\\ \cos \theta = \dfrac{1}{{\sec \theta }} \\\

Replacingsec⁡x\sec x, cot⁡x\cot xandtan⁡x\tan x
1cos⁡xcos⁡xsin⁡x+sin⁡xcos⁡x=sin⁡x ⇒1cos⁡xcos⁡2x+sin⁡2xcos⁡xsin⁡x=sin⁡x  \dfrac{{\dfrac{1}{{\cos x}}}}{{\dfrac{{\cos x}}{{\sin x}} + \dfrac{{\sin x}}{{\cos x}}}} = \sin x \\\ \Rightarrow \dfrac{{\dfrac{1}{{\cos x}}}}{{\dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\cos x\sin x}}}} = \sin x \\\
Now, to solve this question we will take the LHS of the equation and solve it in such a manner that we will get the RHS. If this happens, we can safely conclude that LHS=RHS and hence proved
So, taking the RHS
RHS=1cos⁡xcos⁡2x+sin⁡2xcos⁡xsin⁡xRHS = \dfrac{{\dfrac{1}{{\cos x}}}}{{\dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\cos x\sin x}}}}
We know that,
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
Replacing this in the LHS, we will get:-
RHS=1cos⁡x1cos⁡xsin⁡x RHS=1cos⁡x×cos⁡xsin⁡x1 RHS=sin⁡x1 RHS=sin⁡x=RHS  RHS = \dfrac{{\dfrac{1}{{\cos x}}}}{{\dfrac{1}{{\cos x\sin x}}}} \\\ RHS = \dfrac{1}{{\cos x}} \times \dfrac{{\cos x\sin x}}{1} \\\ RHS = \dfrac{{\sin x}}{1} \\\ RHS = \sin x = RHS \\\
Hence, when we solve the LHS of the equation, we end up with a value which was our RHS. This thing proves that RHS=LHS and hence the question given to us is finally proved.

Note: Here we have remembered the reciprocal of the trigonometric function. Hre in this solution we will use the reciprocal of sin, cos and cot. Properties of trigonometry is one of the most important things in the chapter. You can’t solve any question in this chapter without knowing properties. It is a must if you want to ace every trigonometry question. Therefore, try to remember them all and use them strategically to solve all of your problems.