Question

csc^2(x) + sec^2(x) = csc^2(x)sec^2(x)

Answer 1

what is your question?

Answer 2

prove the trig identity so I have to make the LS = RS

Answer 3

sometimes, we can starting from the right side.
we knowed that
csc = 1/sin and sec = 1/cos

so,
csc^2(x) sec^2(x)
= 1/sin^2(x) * 1/cos^2(x)
= 1/(sin^2x cos^2 x)

remmber this identity :
1 = sin^2x + cos^2x

therefore,
1/(sin^2x cos^2 x)
= (sin^2(x) + cos^2(x))/(sin^2x cos^2 x)
splite into 2 parts, get
= sin^2x/(sin^2x cos^2 x) + cos^2x/(sin^2x cos^2 x)
= 1/cos^2x + 1/sin^2(x)
= sec^2(x) + csc^2(x)

Answer 4

would you always want to split one side if its like A + B / AB? so it would then be
A / AB + B / AB ? did i write that right?

Answer 5

LHS\[\frac{ 1 }{ \sin^2x }+\frac{ 1 }{ \cos^2x }=\frac{ 1 }{ sin^2x* cos^2x }\]
\[\frac{ \sin^2+\cos^2x }{ \sin^2x* \cos^2x}=\frac{ 1 }{ \sin^2x *\cos^2x}\]
\[LHS=\frac{ 1 }{ \sin^2x *\cos^2x }=\frac{ 1 }{ \sin^2x *\cos^2x }=RHS\]

Answer 6

yes, then cancels wich similar terms

Answer 7

* which

Answer 8

alright thanks again!

Answer 9

welcome