The expression \[\tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A) = \]
A. \[0\]
B. \[2\tan A\]
C. \[2\cot A\]
D. \[2(\tan A - \cot A)\]
Answer
528k+ views
Hint: 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 \[\dfrac{\pi }{2}\], i.e. \[\cot ({90^ \circ } + \theta ) = - \tan \theta \] . 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 \[\dfrac{\pi }{2}\] trigonometric ratios of sum of complementary angles does not change. i.e. \[\cot ({180^ \circ } + \theta ) = \cot \theta \] and \[\cot ({360^ \circ } - \theta ) = - \cot \theta \] where \[\theta \] is an angle.
Hence \[\cot ({180^ \circ } + A) = \cot A\] and \[\cot ({360^ \circ } - A) = - \cot A\].
Also we know that at odd multiples of \[\dfrac{\pi }{2}\] .Trigonometric ratios of complementary angle changes from cotangent to tangent and vice-versa . i.e. \[\cot ({90^ \circ } + \theta ) = - \tan \theta \]
Hence, we have \[\cot ({90^ \circ } + A) = - \tan A\].
Now from the given question we have,
\[ \Rightarrow \tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A)\]
Putting the values from above, then
\[ \Rightarrow \tan A + (\cot A) + ( - \tan A) + ( - \cot A)\]
On simplifying,
\[ \Rightarrow \tan A + \cot A - \tan A - \cot A\]
\[ \Rightarrow 0\]
Hence, \[\tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A) = 0\]
Hence, option A is correct.
Note:When the value of \[180 > \theta > 90\] It lies in the second quadrant, where only sine/cosecant is positive and other T-Ratios are negative. Hence we have \[\cot ({90^ \circ } + A) = - \tan A\] Here the negative sign denotes that the values of cotangent in the second quadrant are negative. When the value of \[{180^ \circ } < \theta < {270^ \circ }\] lies in the third quadrant, where only tangent/cotangent is positive and other T-Ratios are negative. Hence we have \[\cot ({180^ \circ } + A) = \cot A\]. When the value of \[{270^ \circ } < \theta < {360^ \circ }\] It lies in the fourth quadrant, where only cosine/secant is positive and other T-Ratios are negative. Hence we have \[\cot ({360^ \circ } - A) = - \cot A\] Here the negative sign denotes that the values of cotangent in the fourth quadrant are negative.
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 \[\dfrac{\pi }{2}\] trigonometric ratios of sum of complementary angles does not change. i.e. \[\cot ({180^ \circ } + \theta ) = \cot \theta \] and \[\cot ({360^ \circ } - \theta ) = - \cot \theta \] where \[\theta \] is an angle.
Hence \[\cot ({180^ \circ } + A) = \cot A\] and \[\cot ({360^ \circ } - A) = - \cot A\].
Also we know that at odd multiples of \[\dfrac{\pi }{2}\] .Trigonometric ratios of complementary angle changes from cotangent to tangent and vice-versa . i.e. \[\cot ({90^ \circ } + \theta ) = - \tan \theta \]
Hence, we have \[\cot ({90^ \circ } + A) = - \tan A\].
Now from the given question we have,
\[ \Rightarrow \tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A)\]
Putting the values from above, then
\[ \Rightarrow \tan A + (\cot A) + ( - \tan A) + ( - \cot A)\]
On simplifying,
\[ \Rightarrow \tan A + \cot A - \tan A - \cot A\]
\[ \Rightarrow 0\]
Hence, \[\tan A + \cot ({180^ \circ } + A) + \cot ({90^ \circ } + A) + \cot ({360^ \circ } - A) = 0\]
Hence, option A is correct.
Note:When the value of \[180 > \theta > 90\] It lies in the second quadrant, where only sine/cosecant is positive and other T-Ratios are negative. Hence we have \[\cot ({90^ \circ } + A) = - \tan A\] Here the negative sign denotes that the values of cotangent in the second quadrant are negative. When the value of \[{180^ \circ } < \theta < {270^ \circ }\] lies in the third quadrant, where only tangent/cotangent is positive and other T-Ratios are negative. Hence we have \[\cot ({180^ \circ } + A) = \cot A\]. When the value of \[{270^ \circ } < \theta < {360^ \circ }\] It lies in the fourth quadrant, where only cosine/secant is positive and other T-Ratios are negative. Hence we have \[\cot ({360^ \circ } - A) = - \cot A\] Here the negative sign denotes that the values of cotangent in the fourth quadrant are negative.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

