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Question: How do you prove the identity \(\dfrac{{\tan x}}{{\csc x + \cot x}} = \sec x - 1\)...

How do you prove the identity tan⁡xcsc⁡x+cot⁡x=sec⁡x−1\dfrac{{\tan x}}{{\csc x + \cot x}} = \sec x - 1

Explanation

Solution

We will begin with the left-hand side of the equation and then we will solve it by first converting all the terms in sin and cos and then we will take the LCM and simplify the expression. After that take the right-hand side and convert all terms in sin and cos. Then equate both sides. We will use the formula sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 which will help us in proving the given expression.

Complete step-by-step answer:
It is mentioned in the question that
⇒tan⁡xcsc⁡x+cot⁡x=sec⁡x−1\Rightarrow \dfrac{{\tan x}}{{\csc x + \cot x}} = \sec x - 1 ….. (1)
Now beginning with the left-hand side of the equation (1) we get,
⇒tan⁡xcsc⁡x+cot⁡x\Rightarrow \dfrac{{\tan x}}{{\csc x + \cot x}}
Now converting cosec, cot, and tan in the above equation in terms of cos and sin and hence we get,
⇒sin⁡xcos⁡x1sin⁡x+cos⁡xsin⁡x\Rightarrow \dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}}}
As the denominator value is the same, add the numerator part,
⇒sin⁡xcos⁡x1+cos⁡xsin⁡x\Rightarrow \dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{1 + \cos x}}{{\sin x}}}}
Simplify the terms,
⇒sin⁡2xcos⁡x(1+cos⁡x)\Rightarrow \dfrac{{{{\sin }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}
Now we know that,
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
So, applying this in the equation we get,
⇒1−cos⁡2xcos⁡x(1+cos⁡x)\Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}
We know that,
(a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)
So, applying this in the equation we get,
⇒(1+cos⁡x)(1−cos⁡x)cos⁡x(1+cos⁡x)\Rightarrow \dfrac{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}{{\cos x\left( {1 + \cos x} \right)}}
Cancel out the common terms,
⇒1−cos⁡xsin⁡x\Rightarrow \dfrac{{1 - \cos x}}{{\sin x}} ….. (2)
Now take the right-hand side of the equation (1) we get,
⇒sec⁡x−1\Rightarrow \sec x - 1
Now converting sec in the above equation in terms of cos and hence we get,
⇒1cos⁡x−1\Rightarrow \dfrac{1}{{\cos x}} - 1
Taking the LCM we get,
⇒1−cos⁡xcos⁡x\Rightarrow \dfrac{{1 - \cos x}}{{\cos x}} ….. (3)
Since from equation (2), we can say that the left-hand side is equal to the right-hand side in equation (3).

Hence, we have proved the given expression.

Note:
In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further but here the key is to substitute 1 in place of sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1. Then we need to divide each term in the numerator separately by the denominator to get the left-hand side equal to the right-hand side of the equation.