Asked by sunny chauhan
___1____ __ ___1_____ = -2tan(x)sec(x)
1-cosecx 1+cosecx
Answers
Answered by
Reiny
Iwill read that as
1/(1 - cscx) - 1/(1 + cscx) = -2tanx secx
LS = (1 + cscx - (1 - cscx))/(1 - csc^2 x)
= 2cscx/(-cot^2 x)
= -2 (1/sinx)(tan^2 x)
= -2(1/sinx)(sin^2 x/cos^2 x)
= -2 cscx/cos^2 x
= -2 cscx sec^2 x
= RS
1/(1 - cscx) - 1/(1 + cscx) = -2tanx secx
LS = (1 + cscx - (1 - cscx))/(1 - csc^2 x)
= 2cscx/(-cot^2 x)
= -2 (1/sinx)(tan^2 x)
= -2(1/sinx)(sin^2 x/cos^2 x)
= -2 cscx/cos^2 x
= -2 cscx sec^2 x
= RS
Answered by
Steve
1/(1-csc) - 1/(1+csc)
= ((1+csc)-(1-csc))/(1-csc^2)
= 2csc/cot^2
= 2csc tan^2
= 2/sin sin^2/cos^2
= 2sin/cos^2
= 2tan sec
I seem to have lost a - sign. Hard to tell just what the original equation was.
= ((1+csc)-(1-csc))/(1-csc^2)
= 2csc/cot^2
= 2csc tan^2
= 2/sin sin^2/cos^2
= 2sin/cos^2
= 2tan sec
I seem to have lost a - sign. Hard to tell just what the original equation was.
Answered by
Steve
Oops. I see where. Reiny did it right.
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