Question

Verify cotxsec^4x=cotx+2tanx+tan^3x.
Verify 1+sec^2xsin^2x=sec^2x
Verify (sinx/1-cosx)+(sinx/1+cosx)=2cscx
Can someone walk me through the steps? I've been working on this for hours and I don't get it!!

Answer 1

\[1 + \sec^2 \theta * \sin^2 \theta = 1 + \frac{ \sin^2 \theta }{ \cos^2 \theta }\]
\[1 + \tan^2 \theta = \sec^2 \theta\]

Answer 2

\[\cot \theta * \sec^4 \theta = \cot \theta * (\sec^2 \theta)\]

Answer 3

\[\cot \theta * \sec^4 \theta = \cot \theta * (\sec^2 \theta )^2 \]
\[\cot \theta * (\sec^2 \theta )^2 =\cot \theta * (1 +\tan^2 \theta )^2 \]
\[\cot \theta * (1 +\tan^2 \theta )^2 = \cot \theta* ( 1 + \tan^4\theta + 2\tan^2\theta)\]
\[\cot \theta (1 + \tan^4\theta + 2 \tan^2\theta ) = \cot \theta + \tan^3\theta + 2\tan \theta\]

Answer 4

3 part
\[\frac{ \sin \theta }{ 1 - \cos \theta }+\frac{ \sin \theta }{ 1 + \cos \theta }=2cscx \]
\[\frac{ \sin \theta }{ 1 - \cos \theta } + \frac{ \sin \theta }{ 1 + \cos \theta } = \frac{ \sin \theta *(1 +\cos \theta) + \sin \theta (1- \cos \theta) }{ (1 - \cos^2 \theta) }\]
\[\frac{ \sin \theta *(1 +\cos \theta) + \sin \theta (1- \cos \theta) }{ (1 - \cos^2 \theta) } = \frac{ \sin \theta + \sin \theta \cos \theta + \sin \theta - \sin \theta \cos \theta }{ \sin^2 \theta }\]

\[\frac{ \sin \theta + \sin \theta \cos \theta + \sin \theta - \sin \theta \cos \theta }{ \sin^2 \theta } = \frac{ 2 \sin \theta }{ \sin^2 \theta }\]

\[\frac{ 2 \sin \theta }{ \sin^2 \theta } = \frac{ 2 }{ \sin \theta } = 2 \csc \theta\]