# Proving trigonometric identity: $\frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x}$ [duplicate]

Prove that $$\frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x}$$

This is my working -

$$\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos x - \cos y \sin x}$$

I used the compound angle relations formula to do this step.

However , now I'm stuck.

I checked out the answer and the answer basically carried on my step by dividing it by $\frac{\cos x \cos y}{\cos x \cos y}$

If I'm not wrong , we cannot add in our own expression right? Because the question is asking to prove left is equal to right. Adding in our own expression to the left hand side will be not answering the question, I feel .

## marked as duplicate by Guy Fsone, Sahiba Arora, user228113, Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 14 '17 at 0:39

• Multiplying the expression by $\frac{cosx cosy}{cosx cosy}$ is like multiplying the expression by 1, and that should not cause any problems. – Shobhit Dec 2 '17 at 9:42
• @user175989 you can set as solved if you are ok – gimusi Dec 3 '17 at 10:00

All you need to do now is$$\frac{\sin y\cos x+\cos y\sin x}{\sin y\cos x-\cos y\sin x}=\frac{\frac{\sin y\cos x+\cos y\sin x}{\cos x\cos y}}{\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}}=\frac{\tan x+\tan y}{\tan x-\tan y}.$$

Consider fractions that are equivalent to $2/3$.

$$\frac{2}{3}\cdot\frac{2}{2} = \frac{4}{6}$$ $$\frac{2}{3}\cdot\frac{3}{3} = \frac{6}{9}$$ $$etc.$$

Conclusion ... you're allowed to multiply by one.

Similarly, with the expression $\displaystyle\frac{\sin y\cos x+\cos y \sin x}{\sin y \cos x−\cos y \sin x}$, you're allowed to multiply by $\displaystyle\frac{\frac{1}{\cos y \cos x}}{\frac{1}{\cos y \cos x}}$ because $$\displaystyle\frac{\frac{1}{\cos y \cos x}}{\frac{1}{\cos y \cos x}} = 1$$

And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows $$\displaystyle\frac{\frac{\sin y\cos x+\cos y \sin x}{1}}{\frac{\sin y \cos x−\cos y \sin x}{1}}$$ before multiplying.

• What does $\cos y \cos x$ do to the expression ? there's nothing that can be cancelled out due to the '+' sign in the expression – user175089 Dec 3 '17 at 9:19
• Try multiplying the two expressions together, as suggested, and think of the distributive property when you do. – John Joy Dec 4 '17 at 14:25

It's ok, let's divide by $\cos y \cos x$

NOTE

In general for $c \neq 0$

$$\frac{a}{b} \iff \frac{\frac{a}{c}}{ \frac{b}{c}}$$

in this case the check is ok because $\tan x$ is not defined when $\cos x=0$.

$$\frac{\tan x+\tan y}{\tan x-\tan y} =\frac{\frac{\sin y\cos x+\cos y\sin x}{\cos x\cos y}}{\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}}= \frac{\sin y\cos x+\cos y\sin x}{\sin y\cos x-\cos y\sin x}=\frac{\sin (y+x)}{\sin (y-x)}$$

which gives the results since we know that $$\sin(x-y) =\sin y\cos x-\cos y\sin x$$and $$\sin(x+y) =\sin y\cos x+\cos y\sin x$$