Question

If \[xy < 1\], \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \_\_\_\_\_\_\_.\]

Answer 1

Hint: We have given an inverse trigonometric expression in \[{\tan ^{ - 1}}\] and we have to find its value. Firstly, we will take the function equal to \[\theta \]. Then, we take the\[\tan \] function on both sides. Then, we will apply the identity of \[\tan \left( {A + B} \right)\]. This will lead us to the required results. The tan function is a trigonometric function. It is a periodic function. The simplest way to understand the tangent function is to use the unit circle.

Complete answer:
We have given inverse trigonometric expressions.
$\Rightarrow$\[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y\]
We have to find its value.
Let \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y\] be equal to some angle \[\theta \].
So, \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \theta \]
Now, taking both sides the \[\tan \] function, we get
$\Rightarrow$\[\tan \left[ {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y} \right] = \tan \theta ........(1)\]
The left-hand side of the equation is in the form \[\tan \left( {A + B} \right)\], where \[A\] is \[{\tan ^{ - 1}}x\] and \[B\] is \[{\tan ^{ - 1}}y\]
Also, we have identity of \[\tan \left( {A + B} \right)\]
$\Rightarrow$\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
So, left hand side of the equation will be
$\Rightarrow$\[\dfrac{{\tan \left( {{{\tan }^{ - 1}}x} \right) + \tan \left( {{{\tan }^{ - 1}}y} \right)}}{{1 - \tan \left( {{{\tan }^{ - 1}}x} \right)\tan \left( {{{\tan }^{ - 1}}y} \right)}}\]
Now, the function \[\tan \] and \[{\tan ^{ - 1}}\] are opposite to each other so will cancel each other’s change. So, the left-hand side becomes
$\Rightarrow$\[\dfrac{{x + y}}{{1 - xy}}\]
So, equation (i) can be written as
$\Rightarrow$\[\tan \theta = \dfrac{{x + y}}{{1 - xy}}\]
Again taking \[\tan \] inverse function on both sides
$\Rightarrow$\[{\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]
Putting value of \[\theta \], we get

\[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]

Note: Trigonometry is the branch of mathematics that studies the relationship between side length and angles of the triangle. Trigonometry has six trigonometric functions which are \[\sin ,\cos, \tan ,\cos ec,\sec \] and \[\cot \]. Trigonometric functions are the real functions which relate an angle of right-angled triangles to the ratio of two sides of a triangle.
Trigonometric functions are also called circular functions. With the help of these trigonometric functions, we can drive lots of trigonometric formulas.