Recent content by PanTh3R

would it be... \frac{\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta} then \frac{1-\cos^2+\cos^2\theta}{\cos\theta\sin\theta} into \frac{1}{\cos\theta\sin\theta} i feel so frustrated :confused: sry

im sorry I am really bad with these i just started doing them um so i do \frac{\sin^2\theta+\cos\theta}{\cos\theta\sin\theta} i still don't get it won't that make it more complicated? :confused:

so i get... \sin\theta\frac{1}{\cos\theta}+cos\theta\frac{1}{\sin\theta} then i put each in one fraction right? \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} to \tan\theta+cot\theta is that the simplest it can get?

Homework Statement simplify: \sin\theta\sec\theta+\cos\theta\csc\theta Homework Equations Reciprocal identities, Quotient identities, Pythagorean identities The Attempt at a Solution \sin\theta\sec\theta+\frac{1}{\sec\theta}\frac{1}{\sin\theta}...

thank you so much for helping me :biggrin:

cos(x)-\frac{cos^{2}(x)}{cos(x)-sin(x)} after long staring and thinking a light bulb just lit in my head lol so... \frac{\cos^2(x)-\sin(x)\cos(x)-\cos^2(x)}{cos(x)-sin(x)} to \frac{-\sin(x)\cos(x)}{cos(x)-sin(x)} then all multiplied by -1 equals...

when i do that won't i get [-cos^2(x)sin(x)]/[cos(x)-sin(x)] then what...sorry I'm not that good as these kinda stuff

Homework Statement Verify the Identity: cos(x)-[cos(x)/1-tan(x)] = [(sin(x)cos(x)]/[sin(x)-cos(x)] b]2. Homework Equations [/b] reciprocal Identities, quotient Identities, Pythagorean Identities [b]3. The Attempt at a Solution cos(x)-[cos(x)/1-tan(x)] =...

oops i put in the problem wrong and i figured it out(the second one) f(x)=1/2x^3-x x(1/2x^2-1)=0 x=0 1/2x^2-1=0 1/2x^2=1 x^2=2 x= +-square root of 2 Thank you for your assistance.

Homework Statement Find the zeros of the function algebraically Homework Equations f(x)=x/(9x^2-4) and f(x)=x^3-x The Attempt at a Solution i was gone for about a month and a half from the class because i had to move and I am kinda rusty ATTEMPT- f(x)=x/(9x^2-4) Multiplied by...