Question

COT ^2 A + CSC ^2 A = - COT^4 A + CSC^4 A

Help Please! trying to prove using trig identities?
help please. thank you

Answer 1

LHS \[=\frac{ 1 }{ \tan^2A }+\frac{ 1 }{ \sin^2A }\]

Answer 2

\[=\frac{ \cos^2A }{ \sin^2A }+\frac{ 1 }{ \sin^2A }\]\[=\frac{ \cos^2A +1 }{ \sin^2A }\]

Answer 3

in the same way you can show :

RHS\[=\frac{ 1-\cos^4A }{ \sin^4A }\]

Answer 4

i'm confused how you came up with the RHS ? sorry :(

Answer 5

doing the same:
cot = cos / sin
csc = 1 / sin

Answer 6

wouldn't the RHS be,
\[= - (\frac{\cos ^{2 }A}{\sin ^{2}A })^{2}\]

Answer 7

RHS
\[-\cot^4A+\csc^4A\]\[=-\frac{ \cos^4A }{ \sin^4A }+\frac{ 1 }{ \sin^4A }\]

Answer 8

So what would be the next step to get the powers down to 2 and get rid of the negative?

Answer 9

1-cos^4A/sin^4 then use a^2-b^2 formula in numerator you get (1+cos^2A )(sin^2A)/sin^4A

Answer 10

u cancel out sin2a and hence LHS = RHS like @PaxPolaris said

Answer 11

RHS\[=\frac{ 1-\cos^4A }{ \sin^4A }\]\[=\frac{ (1+\cos^2A )(1-\cos^2A )}{ \sin^4A }\]

Answer 12

and since \(1-\cos^2A=\sin^2A\) ...

Answer 13

awwww... ok so then 1-cos^2 a becomes sin^2a and cancels out, and the denominator becomes sin^2 a... ok.

Answer 14

awesome. thank you so much.