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Question: Prove that \(\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}\)...

Prove that tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}

Explanation

Solution

Here we have been given an identity and we have to prove that it is true. Firstly we will take the left hand side value and try to get the right hand side value by it. Start by changing the tangent function into sine and cosine functions. Then using the difference identity of sine and cosine expands the values. Finally, multiply or divide by any value so that we get the desired answer.

Complete step-by-step solution:
We have to prove that,
tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}
Firstly we will take the left hand side value,
tan⁡(x−y)\tan \left( x-y \right)….(1)\left( 1 \right)
As we know that
tan⁡A=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A}
Using the above formula in equation (1) we get,
tan⁡(x−y)⇒sin⁡(x−y)cos⁡(x−y)\tan \left( x-y \right)\Rightarrow \dfrac{\sin \left( x-y \right)}{\cos \left( x-y \right)}….(2)\left( 2 \right)
The different identities of sine and cosine are as follows:
sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y
cos⁡(x−y)=cos⁡xcoxy+sin⁡xsin⁡y\cos \left( x-y \right)=\cos xcoxy+\sin x\sin y
Using the above identities in equation (2) we get,
tan⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡ycos⁡xcoxy+sin⁡xsin⁡y\tan \left( x-y \right)=\dfrac{\sin x\cos y-\cos x\sin y}{\cos xcoxy+\sin x\sin y}
Now as we want the right hand side term in tangent function and also the first value of denominator should be 11 so we will divide both numerator and denominator by cos⁡xcos⁡y\cos x\cos y as follows,
tan⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡y+sin⁡xsin⁡ycos⁡xcos⁡y\tan \left( x-y \right)=\dfrac{\dfrac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}}{\dfrac{\cos x\cos y+\sin x\sin y}{\cos x\cos y}}
⇒tan⁡(x−y)=sin⁡xcos⁡ycos⁡xcos⁡y−cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡ycos⁡xcos⁡y+sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow \tan \left( x-y \right)=\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y}-\dfrac{\cos x\sin y}{\cos x\cos y}}{\dfrac{\cos x\cos y}{\cos x\cos y}+\dfrac{\sin x\sin y}{\cos x\cos y}}
Simplifying further we get,
⇒tan⁡(x−y)=sin⁡xcos⁡x−sin⁡ycos⁡y1+sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow \tan \left( x-y \right)=\dfrac{\dfrac{\sin x}{\cos x}-\dfrac{\sin y}{\cos y}}{1+\dfrac{\sin x\sin y}{\cos x\cos y}}
So using tan⁡A=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A} we can rewrite the above value as,
⇒tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\Rightarrow \tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}
So we got the right hand side value.
Hence proved that tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}.

Note: Trigonometry functions are very useful in solving the problem related to right-angle triangle sides and angles. It is a very important branch in the history of mathematics and the concept of trigonometry is given by Greek mathematician Hipparchus. The six basic trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent. In this type of question we should use the relation between the six trigonometric functions so as to expand the values and then by using different identities prove it.