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Mathematics 23 Online
OpenStudy (anonymous):

prove that 1 + 1/tan^2x = 1/sin^2x

OpenStudy (kinggeorge):

\[\begin{align} 1+{1 \over \tan^2(x)} &={1 \over \sin^2(x)}\\{\sin^2(x) \over \sin^2(x)}+{\cos^2(x) \over \sin^2(x)} &={1 \over \sin^2(x)} \\ {\sin^2(x)+cos^2(x) \over \sin^2(x)}&={1 \over \sin^2(x)} \\ {1 \over \sin^2(x)}&={1 \over \sin^2(x)} \end{align}\]So we're done.

OpenStudy (anonymous):

rewrite 1/tan^2x as cos^2x/sin^2x. rewrite 1 as sin^2x/sin^2x so that you have a common denominator. add. sin^2x + cos^2x=1. so both sides of the equation should look the same.

OpenStudy (anonymous):

thankss!

OpenStudy (kinggeorge):

You're welcome.

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