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Question: How do you solve \[\tan \left( x+y \right)=\dfrac{\left( \tan x+\tan y \right)}{1-\tan y\tan x}\]?...

How do you solve tan⁡(x+y)=(tan⁡x+tan⁡y)1−tan⁡ytan⁡x\tan \left( x+y \right)=\dfrac{\left( \tan x+\tan y \right)}{1-\tan y\tan x}?

Explanation

Solution

In this problem, we have to solve the given trigonometric identity. Here we can first take the left-hand side and solve for the right-hand side. We know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, we can use this formula first and then we can use the formula sin⁡(x+y)=sin⁡xcos⁡ycos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y\cos x\sin y and cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y. We can divide the result with cos⁡xcos⁡y\cos x\cos y to get the right-hand side.

Complete step-by-step solution:
Here we have to solve the given trigonometric identity.
tan⁡(x+y)=(tan⁡x+tan⁡y)1−tan⁡ytan⁡x\tan \left( x+y \right)=\dfrac{\left( \tan x+\tan y \right)}{1-\tan y\tan x}
We can now take the left-hand side and solve it for the right-hand side.
We know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}.
We can now write the left-hand side of the given trigonometric identity and write it as,
⇒tan⁡(x+y)=sin⁡(x+y)cos⁡(x+y)\Rightarrow \tan \left( x+y \right)=\dfrac{\sin \left( x+y \right)}{\cos \left( x+y \right)} …….. (1)
We can now simplify the above step.
We know that cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y and sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y.
We can now substitute the above formula in (1), we get
⇒tan⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y\Rightarrow \tan \left( x+y \right)=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}
We can now divide cos⁡xcos⁡y\cos x\cos y on both the numerator and the denominator, we get
⇒tan⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡y−sin⁡xsin⁡ycos⁡xcos⁡y\Rightarrow \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}}
We can now simplify the above step, we get
⇒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}}
We can now cancel similar terms in both the numerator and the denominator in the above step, 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}}
We can now simplify the above step using the formula tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, we get
⇒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}
LHS = RHS.
Therefore, tan⁡(x+y)=(tan⁡x+tan⁡y)1−tan⁡ytan⁡x\tan \left( x+y \right)=\dfrac{\left( \tan x+\tan y \right)}{1-\tan y\tan x}.

Note: We should always remember the trigonometric formula such as tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}. We should also remember the identities sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y and cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y. We should also know to substitute the formulas and simplify it to get the final answer or to solve for both sides.