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

How do you prove the identity tan⁡xcos⁡ecx+cot⁡x=sec⁡x−1\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1 ?

Explanation

Solution

To solve such questions start by separately computing the left-hand side and right-hand side of the given equation. Next, write each term in the given equation to its corresponding sin⁡x\sin x and cos⁡x\cos x values. Then check whether the left-hand side and right-hand side are equal.

Complete step by step answer:
Given tan⁡xcos⁡ecx+cot⁡x=sec⁡x−1\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1
First consider the left-hand side of the given equation, that is,
tan⁡xcos⁡ecx+cot⁡x......(1)\dfrac{{\tan x}}{{\cos ecx + \cot x}}......(1)
We know that
tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}}
Also, it is known that
cos⁡ecx=1sin⁡x\cos ecx = \dfrac{1}{{\sin x}}
And
cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}}
Substitute these in the equation (1)\left( 1 \right) , that is,
sin⁡xcos⁡x1sin⁡x+cos⁡xsin⁡x\dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}}}
Rearranging the terms we get,
sin⁡xcos⁡x(1+cos⁡xsin⁡x)\dfrac{{\sin x}}{{\cos x\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right)}}
Taking sin⁡x\sin x to the numerator we get
sin⁡2xcos⁡x(1+cos⁡x)......(2)\dfrac{{{{\sin }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}......(2)
We also know that,
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
Rearranging this we get,
sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x
Substituting this in the equation (2)\left( 2 \right)
1−cos⁡2xcos⁡x(1+cos⁡x)\dfrac{{1 - {{\cos }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}
We know the identity a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) , so we can write the above equation as,
(1+cos⁡x)(1−cos⁡x)cos⁡x(1+cos⁡x)\dfrac{{\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)}}{{\cos x\left( {1 + \cos x} \right)}}
Canceling out the common terms in the numerator and denominator, we get,
1−cos⁡xcos⁡x......(3)\dfrac{{1 - \cos x}}{{\cos x}}......(3)
Next consider the right-hand side of tan⁡xcos⁡ecx+cot⁡x=sec⁡x−1\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1 , that is
sec⁡x−1\sec x - 1
We know that,
sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}}
Substituting this on the right-hand side, we get,
1cos⁡x−1\dfrac{1}{{\cos x}} - 1
Further simplifying we get,
1−cos⁡xcos⁡x......(4)\dfrac{{1 - \cos x}}{{\cos x}}......(4)
From equation (3)\left( 3 \right) and equation (4)\left( 4 \right) , it can be seen that LHS is equal to RHS.
Therefore, proved the identity tan⁡xcos⁡ecx+cot⁡x=sec⁡x−1\dfrac{{\tan x}}{{\cos ecx + \cot x}} = \sec x - 1 .

Additional information:
All trigonometric ratios of angle can be easily found if one of the trigonometric ratios of an acute angle is given. The value of sin⁡x\sin x or cos⁡x\cos x never goes beyond 11 . Also, the value of sec⁡x\sec x or cos⁡ecx\cos ecx is always greater than or equal to 11 .

Note: In order to solve these types of questions, one should be very clear about the trigonometric identities. The simplest way to solve this type of question is to equate each term of the equation to its corresponding sin⁡x\sin x and cos⁡x\cos x values