Question

(i) If ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}z$ , find z in terms of x and y.
(ii) Express z in terms of x and y, if ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}z$

Answer 1

Hint: We need to express the z in terms of x and y for the given two trigonometric equations. We start to solve the given question using the formulas of ${{\tan }^{-1}}x+{{\tan }^{-1}}y$ , ${{\cos }^{-1}}x+{{\cos }^{-1}}y$ to get the desired result.

Complete step by step solution:
We are given two trigonometric equations in the question and need to express the z in terms of x and y.
We will be solving the given question using the concept and formulae of trigonometry.
(i) According to our question, the first trigonometric equation is given as follows,
$\Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}z$
From trigonometry, we know that
$\Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$
Substituting the same in the above equation, we get,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}z$
Applying tan on both sides of the above equation, we get,
$\Rightarrow \tan \left( {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right)=\tan \left( {{\tan }^{-1}}z \right)$
From trigonometry, we know that the $\tan x$ and ${{\tan }^{-1}}x$ are inverses of each other. The relation between tan and its inverse is given as follows,
$\Rightarrow \tan \left( {{\tan }^{-1}}x \right)=x$
Following the same, we get,
$\Rightarrow \dfrac{x+y}{1-xy}=z$
The above equation can also be written as follows,
$\therefore z=\dfrac{x+y}{1-xy}$
(ii) According to our question, the second trigonometric equation is given as follows,
$\Rightarrow {{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}z$
From trigonometry, we know that
$\Rightarrow {{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)$
Substituting the same in the above equation, we get,
$\Rightarrow {{\cos }^{-1}}\left( xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)={{\cos }^{-1}}z$
Applying cos on both sides of the above equation, we get,
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right) \right)=\cos \left( {{\cos }^{-1}}z \right)$
From trigonometry, we know that the $\cos x$ and ${{\cos }^{-1}}x$ are inverses of each other. The relation between cos and its inverse is given as follows,
$\Rightarrow \cos \left( {{\cos }^{-1}}x \right)=x$
Following the same, we get,
$\Rightarrow xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}=z$
The above equation can also be written as follows,
$\therefore z=xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}$

Note: We must remember that the value of ${{\tan }^{-1}}x$ is not the same as $\dfrac{1}{\tan x}$ and the value of ${{\cos }^{-1}}x$ is not the same as $\dfrac{1}{\cos x}$ . We must know the basic formulae of inverse trigonometry to solve the given question in less time.