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Question: The expression \[\tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } -...

The expression tanA+cot(180+A)+cot(90+A)+cot(360A)=\tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A) =
A. 00
B. 2tanA2\tan A
C. 2cotA2\cot A
D. 2(tanAcotA)2(\tan A - \cot A)

Explanation

Solution

In the given question, we have to find the sum of given trigonometric ratios of sum of angle. We first find the values of each term. We will use the fact that cotangent changes to tangent when it is odd multiple of π2\dfrac{\pi }{2}, i.e. cot(90+θ)=tanθ\cot ({90^ \circ } + \theta ) = - \tan \theta . We will then put the values and sum to get the desired result.

Complete step by step answer:
Given question is based on the sum of trigonometric ratios of the sum of angles.Trigonometric ratios are the ratios of sides of a right angled triangle. The six trigonometric ratios are – sine, cosine, tangent, cotangent, secant and cosecant.Considering the given question, we know that at even multiples of π2\dfrac{\pi }{2} trigonometric ratios of sum of complementary angles does not change. i.e. cot(180+θ)=cotθ\cot ({180^ \circ } + \theta ) = \cot \theta and cot(360θ)=cotθ\cot ({360^ \circ } - \theta ) = - \cot \theta where θ\theta is an angle.
Hence cot(180+A)=cotA\cot ({180^ \circ } + A) = \cot A and cot(360A)=cotA\cot ({360^ \circ } - A) = - \cot A.

Also we know that at odd multiples of π2\dfrac{\pi }{2} .Trigonometric ratios of complementary angle changes from cotangent to tangent and vice-versa . i.e. cot(90+θ)=tanθ\cot ({90^ \circ } + \theta ) = - \tan \theta
Hence, we have cot(90+A)=tanA\cot ({90^ \circ } + A) = - \tan A.
Now from the given question we have,
tanA+cot(180+A)+cot(90+A)+cot(360A)\Rightarrow \tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A)
Putting the values from above, then
tanA+(cotA)+(tanA)+(cotA)\Rightarrow \tan A + (\cot A) + ( - \tan A) + ( - \cot A)
On simplifying,
tanA+cotAtanAcotA\Rightarrow \tan A + \cot A - \tan A - \cot A
0\Rightarrow 0
Hence, tanA+cot(180+A)+cot(90+A)+cot(360A)=0\tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A) = 0

Hence, option A is correct.

Note: When the value of 180>θ>90180 > \theta > 90 It lies in the second quadrant, where only sine/cosecant is positive and other T-Ratios are negative. Hence we have cot(90+A)=tanA\cot ({90^ \circ } + A) = - \tan A Here the negative sign denotes that the values of cotangent in the second quadrant are negative. When the value of 180<θ<270{180^ \circ } < \theta < {270^ \circ } lies in the third quadrant, where only tangent/cotangent is positive and other T-Ratios are negative. Hence we have cot(180+A)=cotA\cot ({180^ \circ } + A) = \cot A. When the value of 270<θ<360{270^ \circ } < \theta < {360^ \circ } It lies in the fourth quadrant, where only cosine/secant is positive and other T-Ratios are negative. Hence we have cot(360A)=cotA\cot ({360^ \circ } - A) = - \cot A Here the negative sign denotes that the values of cotangent in the fourth quadrant are negative.