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Question: (i) If \({{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}z\) , find z in terms of x and y. (ii) Expr...

(i) If tan⁡−1x+tan⁡−1y=tan⁡−1z{{\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⁡−1x+cos⁡−1y=cos⁡−1z{{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}z

Explanation

Solution

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⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y , cos⁡−1x+cos⁡−1y{{\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,
⇒tan⁡−1x+tan⁡−1y=tan⁡−1z\Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}z
From trigonometry, we know that
⇒tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\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,
⇒tan⁡−1(x+y1−xy)=tan⁡−1z\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}z
Applying tan on both sides of the above equation, we get,
⇒tan⁡(tan⁡−1(x+y1−xy))=tan⁡(tan⁡−1z)\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\tan x and tan⁡−1x{{\tan }^{-1}}x are inverses of each other. The relation between tan and its inverse is given as follows,
⇒tan⁡(tan⁡−1x)=x\Rightarrow \tan \left( {{\tan }^{-1}}x \right)=x
Following the same, we get,
⇒x+y1−xy=z\Rightarrow \dfrac{x+y}{1-xy}=z
The above equation can also be written as follows,
∴z=x+y1−xy\therefore z=\dfrac{x+y}{1-xy}
(ii) According to our question, the second trigonometric equation is given as follows,
⇒cos⁡−1x+cos⁡−1y=cos⁡−1z\Rightarrow {{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}z
From trigonometry, we know that
⇒cos⁡−1x+cos⁡−1y=cos⁡−1(xy−(1−x2)(1−y2))\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,
⇒cos⁡−1(xy−(1−x2)(1−y2))=cos⁡−1z\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,
⇒cos⁡(cos⁡−1(xy−(1−x2)(1−y2)))=cos⁡(cos⁡−1z)\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\cos x and cos⁡−1x{{\cos }^{-1}}x are inverses of each other. The relation between cos and its inverse is given as follows,
⇒cos⁡(cos⁡−1x)=x\Rightarrow \cos \left( {{\cos }^{-1}}x \right)=x
Following the same, we get,
⇒xy−(1−x2)(1−y2)=z\Rightarrow xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}=z
The above equation can also be written as follows,
∴z=xy−(1−x2)(1−y2)\therefore z=xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}

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