Asked by Andrea
Verify that each of the following is an identity.
tan^2x-sin^2x=tan^2xsin^2x
I can get it down to cos^2 on the right, but cannot get it to work out on the left.
secx/cosx - tanx/cotx=1
On the left I got down to 1-tan^2, but that clearly doesn't equal 1....
1-2cos^2x/sinxcosx=tanx-cotx
I'm really not sure where to go with this one. Any help would be appreciated. Thanks in advance.
tan^2x-sin^2x=tan^2xsin^2x
I can get it down to cos^2 on the right, but cannot get it to work out on the left.
secx/cosx - tanx/cotx=1
On the left I got down to 1-tan^2, but that clearly doesn't equal 1....
1-2cos^2x/sinxcosx=tanx-cotx
I'm really not sure where to go with this one. Any help would be appreciated. Thanks in advance.
Answers
Answered by
Reiny
For the first, I am going to start on the left
LS = sin^2 x/cos^2 x - sin^2 x
= (sin^2 x - sin^2 xcos^2 x)/sin^2 x
= sin^2x(1 - cos^2 x)/sin^2 x
= (sin^2 x)(sin^2 x)/cos^2 x
= tan^2 x sin^2 x
= RS
for the second
LS = 1/cos^2 x - (sinx)/cosx)]/[cosx/sinx)]
= 1/cos^2 x - sin^2 x/cos%2 x
= (1 - sin^2 x)/cos^2 x
= cos^2 x/cos^2 x
= 1
= RS
LS = sin^2 x/cos^2 x - sin^2 x
= (sin^2 x - sin^2 xcos^2 x)/sin^2 x
= sin^2x(1 - cos^2 x)/sin^2 x
= (sin^2 x)(sin^2 x)/cos^2 x
= tan^2 x sin^2 x
= RS
for the second
LS = 1/cos^2 x - (sinx)/cosx)]/[cosx/sinx)]
= 1/cos^2 x - sin^2 x/cos%2 x
= (1 - sin^2 x)/cos^2 x
= cos^2 x/cos^2 x
= 1
= RS
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