Question

Is the following statement true or false?
\[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2},x \in R\]
A) True.
B) False.

Answer 1

Hint:
In trigonometry the terms tan and cot stand for tangent and cotangent respectively. Inverse trigonometric functions are the inverse functions of the basic trigonometric functions. They are used to get the angle with any trigonometric ratio. We have some formulas in trigonometry which relate these two ratios. One of them is \[\tan \theta = \cot \left( {\dfrac{\pi }{2} - \theta } \right)\]. Using this relation the given problem can be evaluated.

Complete step by step solution:
Tan or cot functions stand for tangent and cotangent of an angle respectively in trigonometry. The tangent of an angle is also defined by the ratio of the sin of the angle to the cosine of the angle, while the cotangent is the ratio of the cosine of the angle to the sine of the angle.
Now we know the relation: \[\tan \theta = \cot \left( {\dfrac{\pi }{2} - \theta } \right)\].
Let \[{\tan ^{ - 1}}x = A\]…..(1)
\[ \Rightarrow \tan A = x\]
Now, \[\tan A = \cot \left( {\dfrac{\pi }{2} - A} \right)\]
\[ \Rightarrow \cot \left( {\dfrac{\pi }{2} - A} \right) = x\]
\[ \Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - A\]…….(2)
Now adding equations (1) and (2),
\[ \Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}x = A + \dfrac{\pi }{2} - A\]
\[ \Rightarrow {\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}\]
Hence, \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}\], \[x \in R\].

Hence the statement is true. Correct option is (a).

Note:
Inverse trigonometric functions are also known as arc functions, i.e. \[{\tan ^{ - 1}}\theta \] can also be represented as \[\arctan \theta \]. The relation we obtained above must be memorised by students and applied in various trigonometric problems. There are also two other results of the same form, involving other angles which also needs to be memorized, they are:
\[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
\[\cos e{c^{ - 1}}x + {\sec ^{ - 1}}x = \dfrac{\pi }{2}\]