Solveeit Logo
Login

Question

Question: If we are given the inverse trigonometric expression as \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y\] \[ =...

If we are given the inverse trigonometric expression as tan⁡−1x+tan⁡−1y{\tan ^{ - 1}}x + {\tan ^{ - 1}}y
=tan⁡−1x+y1−xy,xy<1= {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}},xy < 1
=π+tan⁡−1x+y1−xy,xy>1= \pi + {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}},xy > 1
Evaluate: tan⁡−13sin⁡2α5+3cos⁡2α+tan⁡−1(tan⁡α4){\tan ^{ - 1}}\dfrac{{3\sin 2\alpha }}{{5 + 3\cos 2\alpha }} + {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha }}{4}} \right)
where −π2<α<π2 - \dfrac{\pi }{2} < \alpha < \dfrac{\pi }{2}
A). α\alpha
B). 2α2\alpha
C). 3α3\alpha
D). 4α4\alpha

Explanation

Solution

In the given question, we have been given an expression involving the inverse of a trigonometric expression. We have to solve for an argument of the expression where the argument has some restraints on it. To solve it, we are going to first simplify the expression inside the inverse bracket into the form with which they can directly come out of the brackets. Then we are going to use the standard results and solve for our answer.

Complete step by step solution:
We have to evaluate the value of A=tan⁡−13sin⁡2α5+3cos⁡2α+tan⁡−1(tan⁡α4)A = {\tan ^{ - 1}}\dfrac{{3\sin 2\alpha }}{{5 + 3\cos 2\alpha }} + {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha }}{4}} \right).
Let x=3sin⁡2α5+3cos⁡2αx = \dfrac{{3\sin 2\alpha }}{{5 + 3\cos 2\alpha }}.
First, we are going to simplify the value of xx.
We know, sin⁡2θ=2tan⁡θ1+tan⁡2θ\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} and cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}.
So, x=3×2tan⁡α1+tan⁡2α5+3×1−tan⁡2α1+tan⁡2α=6tan⁡α5(1+tan⁡2α)+3(1−tan⁡2α)x = \dfrac{{3 \times \dfrac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }}}}{{5 + 3 \times \dfrac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}}} = \dfrac{{6\tan \alpha }}{{5\left( {1 + {{\tan }^2}\alpha } \right) + 3\left( {1 - {{\tan }^2}\alpha } \right)}}
Opening the brackets and simplifying,
x=6tan⁡α5+5tan⁡2α+3−3tan⁡2α=6tan⁡α8+2tan⁡2α=3tan⁡α4+tan⁡2αx = \dfrac{{6\tan \alpha }}{{5 + 5{{\tan }^2}\alpha + 3 - 3{{\tan }^2}\alpha }} = \dfrac{{6\tan \alpha }}{{8 + 2{{\tan }^2}\alpha }} = \dfrac{{3\tan \alpha }}{{4 + {{\tan }^2}\alpha }}
Now, x=34tan⁡α1+14tan⁡2α=tan⁡α−14tan⁡α1+14tan⁡α×tan⁡αx = \dfrac{{\dfrac{3}{4}\tan \alpha }}{{1 + \dfrac{1}{4}{{\tan }^2}\alpha }} = \dfrac{{\tan \alpha - \dfrac{1}{4}\tan \alpha }}{{1 + \dfrac{1}{4}\tan \alpha \times \tan \alpha }}
So, tan⁡−1x=tan⁡−1(tan⁡α−14tan⁡α1+14tan⁡α×tan⁡α){\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha - \dfrac{1}{4}\tan \alpha }}{{1 + \dfrac{1}{4}\tan \alpha \times \tan \alpha }}} \right)
Now, it has been given that,
tan⁡−1x+tan⁡−1y{\tan ^{ - 1}}x + {\tan ^{ - 1}}y
=tan⁡−1x+y1−xy,xy<1= {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}},xy < 1
=π+tan⁡−1x+y1−xy,xy>1= \pi + {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}},xy > 1
So, tan⁡−1(tan⁡α−14tan⁡α1+14tan⁡α×tan⁡α)=tan⁡−1(tan⁡(α))−tan⁡−1(tan⁡α4)=α−tan⁡−1(tan⁡α4){\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha - \dfrac{1}{4}\tan \alpha }}{{1 + \dfrac{1}{4}\tan \alpha \times \tan \alpha }}} \right) = {\tan ^{ - 1}}\left( {\tan \left( \alpha \right)} \right) - {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha }}{4}} \right) = \alpha - {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha }}{4}} \right)
Now, we have,
A=α−tan⁡−1(tan⁡α4)+tan⁡−1(tan⁡α4)=αA = \alpha - {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha }}{4}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{\tan \alpha }}{4}} \right) = \alpha
Hence, the correct option is A.

Note: In the given question, we had to simplify an inverse expression of trigonometry. We solved it by simplifying the expression which was the argument of the inverse expression. So, to solve that, we must know the formulae and their results. Then we just used the standard results and solved for our answer.