1. Trig identities help

Guys, i swear im going to fail functions this year T_T

I've been paying attention in class, but i dont get what the teacher is saying at all.

Were supposed to prove that these 2 things are equal.
Can someone pleeease show me how to do this?
because ive tried mixing them around but its not working at all.

so heres the question i have to prove

1 - (sin^2 y)/(1+(1/tan y)) - cos^2 y/1+tan y = sin y cos y

the teacher also wrote

recall: tan y = sin y/cos y

sin^2 y + cos ^2 y = 1

id really appreciate some help T_T thanks so much guys

2. Originally Posted by gazoline
1 - (sin^2 y)/(1+(1/tan y)) - cos^2 y/1+tan y = sin y cos y
I am going to assume that you meant

$\displaystyle 1-\frac{\sin^2y}{1+1/\tan y}-\frac{\cos^2y}{1+\tan y}=\sin y\cos y$

(with the last $\displaystyle \tan y$ included in the denominator).

Begin by combining the fractions on the left side.

$\displaystyle 1-\frac{\sin^2y}{1+1/\tan y}-\frac{\cos^2y}{1+\tan y}$

$\displaystyle =\frac{1+\tan y}{1+\tan y}-\frac{\tan y\sin^2y}{\tan y\,(1+1/\tan y)}-\frac{\cos^2y}{1+\tan y}$

$\displaystyle =\frac{1+\tan y}{1+\tan y}-\frac{\tan y\sin^2y}{1+\tan y}-\frac{\cos^2y}{1+\tan y}$

$\displaystyle =\frac{1+\tan y-\tan y\sin^2y-\cos^2y}{1+\tan y}$

Factor,

$\displaystyle =\frac{1+\tan y\left(1-\sin^2y\right)-\cos^2y}{1+\tan y}$

$\displaystyle =\frac{1+\tan y\cos^2y-\cos^2y}{1+\tan y}$

Convert to sines and cosines.

$\displaystyle =\frac{1+\sin y\cos y-\cos^2y}{1+(\sin y/\cos y)}$

$\displaystyle =\frac{\sin^2y+\sin y\cos y}{(\sin y+\cos y)/\cos y}$

Now just simplify and reduce the fraction, and you're there.

3. thx so much for the quick response i really aprpeciate it