Question

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

Answer 1

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.