Question

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

Answer 1

Hint: 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 \[\tan x=\dfrac{\sin x}{\cos x}\], we can use this formula first and then we can use the formula \[\sin \left( x+y \right)=\sin x\cos y\cos x\sin y\] and \[\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y\]. We can divide the result with \[\cos x\cos y\] to get the right-hand side.

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

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