Maths Help!!! (1 Viewer)

[atar90plus]

Hello
Could you guys please help me with these questions relating to trigonometry more specifically identities. Please show full work and solutions.

Simplify

Q2. c) sec x cot x
i) 5cot^2 x + 5
l) cot x -cotx cos^2 x

Prove that

Q3. a) cos^2 x-1 = -sin^2x
c) 3+3tan^2 x = 3 / 1-sin^2 x
d) sec^2 x -tan^2 x = cosec^2x -cot^2x
g) cos^2 (90-x)cotx = sinx cosx

 

[deswa1]

For 2c)
(secx)(cotx)=(1/cosx)(cosx/sinx)=1/sinx

For 2i) 5(cot^2x+1)=5(cosec^2x) (on using cot^2x+1=cosec^2x)

For 2l) cotx(1-cos^2x)=cotx(sinx)=(cosx/sinx)(sinx)=cosx

Sorry about the lack of latex- I don't have much time now

 

[EpikHigh]

 

[theind1996]

For 2c)
(secx)(cotx)=(1/cosx)(cosx/sinx)=1/sinx

For 2i) 5(cot^2x+1)=5(cosec^2x) (on using cot^2x+1=cosec^2x)

For 2l) cotx(1-cos^2x)=cotx(sinx)=(cosx/sinx)(sinx)=cosx

Sorry about the lack of latex- I don't have much time now

Lol that could be interpreted so differently.

 

[qwerty44]

 

[EpikHigh]

Nicely done I just did that question haha, I need to learn how to use LaTeX :L

 

[deswa1]

For d, both LHS and RHS are equal to one (using cosec^2-cot^2 and sec^2-tan^2).

For g) cos^2(90-x)cotx=sin^2xcotx=(sinx)(sinx)(cosx)/(sinx)=(sinx)(cosx) -> The trick with this one is cos(90-x)=sinx

 

[Carrotsticks]

Moved to Mathematics.

 

[qwerty44]

EDIT: Beaten again D:

 

[gr_111]

Nicely done I just did that question haha, I need to learn how to use LaTeX :L

Have a look at http://www.codecogs.com/latex/eqneditor.php. You can start by using the buttons to work out what does what, and you can either save formula as a pic and attach or copy the code from the bottom and paste into these forums.