Question

Simplify using common denominator:
a) (sin^3theta/cos theta) - (cos^3/sin theta)
b) (cos^2theta/tan theta) - (sin^2theta/cot theta)
c) (sec theta/cos theta) + (cosec theta/sin theta)
d) [(costheta . sintheta)/cot theta] + [(costheta . sintheta)/tan theta]

Answer 1

Srry for a really long question

Answer 2

Add a medal if you want me to draw it

Answer 3

...You mean you want us to give you a medal so you add more to your question? You shouldnt be asking questions to receive medals...

Answer 4

\[ \frac{\sin^3(x)}{\cos(x)} - \frac{\cos^3(x)}{\sin(x)}\]

Answer 5

like that?

Answer 6

Im srry @Confusionist i am new to this 'openstudy' thing. :)

Answer 7

@satellite73 how do you actually do the question

Answer 8

Nah, it's cool! You actually give a medal to the person that you think gave the best response by clicking 'best response' next to their name. :) Why don't you give it a try? I can help you out if you need anything else on the site. If you really like someone's answer you can become their fan and even give them a testimonial.

Answer 9

Alright thanks @Confusionist

Answer 10

This is really 4 questions, right?

Answer 11

Yes... im sorry for bothering all of you!

Answer 12

Is the first one what Satellite posted?

Answer 13

take them 1 at a time

Answer 14

Is the first one what Satellite posted?

Answer 15

yes but i dont know how to finish it..

Answer 16

The instructions say to get a common denominator which would be sin(x)cos(x)

Answer 17

\[\frac{\sin^3x}{\cos x}\times \frac{\sin x}{\sin x}-\frac{\cos ^3x}{\sin x}\times\frac{\cos x}{\cos x}\]

Answer 18

\[\frac{\sin ^4x-\cos ^4x}{\sin x \cos }=\frac{(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2x)}{\sin x \cos x}=\frac{\sin ^2x-\cos ^2x}{\sin x \cos x}\]

Answer 19

Wow Thanks! Very nice of you to write this all down! :)

Answer 20

Now you try the next one.

Answer 21

ok thats right. I should learn from this

Answer 22

btw srry if i dont know/clueless to all this work because this is extension for me (im in yr 10 in a school in Australia)

Answer 23

\[\frac{\cos ^2x}{\tan x}-\frac{\sin ^2x}{\cot x}=\frac{\cos ^2x}{\tan x}\times\frac{\cot x}{\cot x}-\frac{\sin ^2x}{\cot x}\times\frac{\tan x}{\tan x}=\]

Answer 24

\[\frac{\cos ^2x \cot x}{\tan x \cot x}-\frac{\sin ^2x \tan x}{\tan x \cot x}=\frac{\cos ^2x \cot x-\sin ^2xtanx}{\tan x \cot x}\]

Answer 25

\[\cos ^2x \times\frac{\cos x}{\sin x }-\sin ^2x \times\frac{\sin x}{\cos x}=\frac{\cos ^3x}{\sin x}-\frac{\sin ^3x}{\cos x}\]

Answer 26

\[\frac{\cos ^4x-\sin ^4x}{\sin x \cos x}\]

Answer 27

\[\frac{(\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)}{\sin x \cos x}= \frac{\cos ^2x-\sin ^2x}{\sin x \cos x}\]