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Question: Find the value of \({\tan ^{ - 1}}5 + {\tan ^{ - 1}}3 - {\cot ^{ - 1}}\left( {\dfrac{4}{7}} \right) ...

Find the value of tan⁡−15+tan⁡−13−cot⁡−1(47){\tan ^{ - 1}}5 + {\tan ^{ - 1}}3 - {\cot ^{ - 1}}\left( {\dfrac{4}{7}} \right) .
A) π3\dfrac{\pi }{3}
B) π4\dfrac{\pi }{4}
C) π2\dfrac{\pi }{2}
D) π6\dfrac{\pi }{6}

Explanation

Solution

We have to solve the question involving the use of the inverse trigonometric function tan⁡−1{\tan ^{ - 1}} and cot⁡−1{\cot ^{ - 1}}. We will first add the first two terms given in the expression we will evaluate them using the standard formula and then solve it further. Also remember that the inverse functions of tan and cot are complementary to each other.

Formula used:
The formula for adding two tan⁡−1{\tan ^{ - 1}} function is given as:
tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}}

Complete step by step answer:
We are given the expression:
tan⁡−15+tan⁡−13−cot⁡−1(47){\tan ^{ - 1}}5 + {\tan ^{ - 1}}3 - {\cot ^{ - 1}}\left( {\dfrac{4}{7}} \right)
We will first add the first two terms using the standard formula and then solve it further.
⇒tan⁡−1  [5 + 3][1 − 5 × 3]  − cot−1  47\Rightarrow {\tan ^{ - 1}}\;\dfrac{{\left[ {5{\text{ }} + {\text{ }}3} \right]}}{{\left[ {1{\text{ }}-{\text{ }}5{\text{ }} \times {\text{ }}3} \right]{\text{ }}}}{\text{ }}-{\text{ co}}{{\text{t}}^{ - 1}}\;\dfrac{4}{7}
Solving the expression further we get,
⇒tan⁡−1  (8− 14 ) − cot−1  47\Rightarrow {\tan ^{ - 1}}\;\left( {\dfrac{8}{{-{\text{ }}14}}{\text{ }}} \right){\text{ }}-{\text{ co}}{{\text{t}}^{ - 1}}\;\dfrac{4}{7}
⇒tan⁡−1  (−4 7 ) − cot−1  47\Rightarrow {\tan ^{ - 1}}\;\left( { - \dfrac{4}{{{\text{ 7}}}}{\text{ }}} \right){\text{ }}-{\text{ co}}{{\text{t}}^{ - 1}}\;\dfrac{4}{7}
Now since tan⁡−1{\tan ^{ - 1}} is in negative we will use the conversion formula give below to solve it:
tan⁡−1(−a)=π−tan⁡−1a{\tan ^{ - 1}}( - a) = \pi - {\tan ^{ - 1}}a
Using this we get:
⇒π−tan⁡−1  (4 7 ) − cot−1  47\Rightarrow \pi - {\tan ^{ - 1}}\;\left( {\dfrac{4}{{{\text{ 7}}}}{\text{ }}} \right){\text{ }}-{\text{ co}}{{\text{t}}^{ - 1}}\;\dfrac{4}{7}
⇒π−[tan⁡−1  47+cot⁡−1  47]\Rightarrow \pi - \left[ {{{\tan }^{ - 1}}\;\dfrac{4}{7} + {{\cot }^{ - 1}}\;\dfrac{4}{7}} \right]
we know that tan⁡−1{\tan ^{ - 1}} and cot⁡−1{\cot ^{ - 1}} are complementary to each other from the basic premise that :
tan⁡θ=tan⁡(90o−θ)\tan \theta = \tan ({90^o} - \theta )
Hence tan⁡−1a+cot⁡−1a{\tan ^{ - 1}}a + {\cot ^{ - 1}}a will be equal to π2\dfrac{\pi }{2}
So we will write as :
⇒π−π2\Rightarrow \pi - \dfrac{\pi }{2}
⇒π2\Rightarrow \dfrac{\pi }{2}
Therefore, tan⁡−15+tan⁡−13−cot⁡−1(47)=π2{\tan ^{ - 1}}5 + {\tan ^{ - 1}}3 - {\cot ^{ - 1}}\left( {\dfrac{4}{7}} \right) =\dfrac{\pi }{2} option (C) is correct.

Note:
The alternate method to solve this type of question would have been to straightaway convert the cot⁡−1{\cot ^{ - 1}}function given in the expression to the tan⁡−1{\tan ^{ - 1}}function, which will be again be done using the fact that they are complimentary. Thus the formula for the conversion of the cot inverse function to the tan inverse function will be :
cot⁡−1a=π2−tan⁡−1a{\cot ^{ - 1}}a = \dfrac{\pi }{2} - {\tan ^{ - 1}}a
In this case the first few steps would have been same except that in the last steps the answer would have come out such that the term after converting the cot inverse would have been cancelled and only π2\dfrac{\pi }{2} will be left as the correct answer.