Question

How do you verify $ \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} $ ?

Answer 1

Hint: Here, you are given an equation in which the left hand side of the equation contains the trigonometric ratio sine and cosine whereas the right hand side of the equation contains the trigonometric ratio tangent. You need to use the trigonometric identities, such as the expansion of sine and cosine of sum of two angles, that is $ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B $ , $ \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B $ , $ \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} $ and $ \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} $ There are a number of methods in which you can achieve the following identity. One is to directly identify the form of the equation and write it in a compact form, other is that you multiply and divide by a certain term or take common terms or terms out of the equation and then see what is reduced to.

Complete step by step solution:
First, let us identify the form of the numerator and denominator. Now, what we will consider is the sine of the sum of two angles and also the cosine of the sum of two angles.
For sine, we have
$ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B $ and that for cosine, we have
$ \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B $ .
If you consider $ A = x $ and $ B = y $ , you can see that the numerator and denominator of the left-hand side of the given equation shows resemblance. So, we have
$ \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 $ . Now, we will substitute this in our equation. So, we will have
$ \dfrac{{\sin \left( {x + y} \right)}}{{\cos \left( {x + y} \right)}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} $ .
Now, you know that tangent of an angle is equal to sine of that angle divided by the cosine of that angle, mathematically, $ \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} $ . If $ \theta = x + y $ , we get $ \dfrac{{\sin \left( {x + y} \right)}}{{\cos \left( {x + y} \right)}} = \tan \left( {x + y} \right) $ . Now, the tangent of sum of two angles is given by $ \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} $ , again, let $ A = x $ and $ B = y $ , we get $ \tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} $ . This was our first approach to verify the identity.
Now, let us go for another method.
The left-hand side is
 $ \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} $ , what you do is take $ \cos x\cos y $ common from both numerator and denominator. You get
 \[\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\dfrac{{\sin x\cos y}}{{\cos x\cos y}} + \dfrac{{\cos x\sin y}}{{\cos x\cos y}}}}{{\dfrac{{\cos x\cos y}}{{\cos x\cos y}} - \dfrac{{\sin x\sin y}}{{\cos x\cos y}}}}\]
Use $ \tan x = \dfrac{{\sin x}}{{\cos x}} $ and $ \tan y = \dfrac{{\sin y}}{{\cos y}} $ , we get \[\dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}\] . Hence proved.
Therefore, $ \dfrac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y - \sin x\sin y}} = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}} $ , identity verified.

Note: The second way is actually preferred. If you do not remember the expansion of $ \tan \left( {A + B} \right) $ , this will help you. It is also a proof of expansion of $ \tan \left( {A + B} \right) $ . You need to memorize the trigonometric identities such as $ \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B $ , $ \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B $ , $ \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} $ and $ \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} $ . Also, the trick of taking common out or multiplying and dividing a certain term helps you to write the equation in a compact format, so, keep this in mind.