i need help asap

[itz_jenny]

i cant figure these questions out.....please try to help me wit all of them but its ok if u only get a couple


sin t csc t = 1


cot^2 y (sec^2 y - 1) = 1


cos^2 B - sin^2 B = 1 - 2 sin^2 B


cot^3 t/csc t = cos t (csc^2 t - 1)


sec^6x(sec x tan x) - sec^4x( sec x tan x) = sec^5 x tan^3 x


sec x - cos x = sin x tan x


5/tan x - sec x



PLZ help me out with these problems....

 

[stapel]

Are you solving equations? Providing identities? Some of each?

Please reply with the instructions for each of the seven exercises, showing what you have tried thus far (so the tutors can see where you're needing help).

Thank you.

Eliz.

 

[soroban]

Hello, itz_jenny!

i cant figure these questions out

Do you know anything about trigonometry?

\(\displaystyle \sin t\cdot\csc t\;=\;1\)

You should know that: \(\displaystyle \,\csc t\:=\:\frac{1}{\sin t}\)

So you have: \(\displaystyle \,\sin t\cdot\csc t\;=\;\sout{\sin t}\,\cdot\,\frac{1}{\sout{\sin t}}\;=\;1\)

It doesn't get any simpler . . .

\(\displaystyle \cot^2 y (\sec^2 y\,-\,1)\;=\;1\)

You should know that: \(\displaystyle \,\cot y \,=\,\frac{1}{\tan y}\;\) and \(\displaystyle \;\sec^2y\,-\,1\:=\:\tan^2y\)

So you have: \(\displaystyle \;\cot^2y(\sec^2y\,-\,1)\;=\;\frac{1}{\sout{\tan^2y}}\,\cdot\,\sout{\tan^2y} \;=\; 1\)

\(\displaystyle \cos^2B\,-\,\sin^2 B\;=\;1\,-\,2\sin^2B\)

You should know that: \(\displaystyle \,\sin^2B\,+\,\cos^2B\:=\:1\;\;\Rightarrow\;\;\cos^2B\:=\:1\,-\,\sin^2B\)

So you have: \(\displaystyle \,\cos^2B\,-\,\sin^2B\;=\;(1\,-\,\sin^2B)\,-\,\sin^2B\;=\;1\,-\,2\sin^2B\)


Are you getting the idea?