Solving an Equation Involving Trigonometric Functions

[tornzaer]

Homework Statement

(1-cos^2x)(1+tan^2x) = tan^2x

Homework Equations

N/A

The Attempt at a Solution

(1-cos^2x)(1+tan^2x) = tan^2x
L.S.
= (sin^2x)(1+sin^2x/cos^2x)
= sin^2x+(sin^4x/cos^2x)

Now, I get a common denominator, but it's not doing anything for me. Did I do the right thing in converting the tan?

 

[rocomath]

Mess with the left side ... what is 1-cos^(2)x and convert tan^(2)x into sine/cosine.

 

[tornzaer]

Mess with the left side ... what is 1-cos^(2)x and convert tan^(2)x into sine/cosine.

Already did that on the first post. See, I missed all the lessons at school so I'm trying to piece everything in by myself. Can you please take a look at what I did and specify the next steps?

 

[rocomath]

Now, I get a common denominator, but it's not doing anything for me.

You did not do this step correctly, do it again and you will get an identity that simplifies everything.

 

[tornzaer]

What I'm getting is below.

= sin^2x+(sin^4x/cos^2x)
= ((sin^2x)(cos^x)+sin^4x)/cos^2x

The cos^2x is the common denominator.

 

[rocomath]

\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}

\sin^{2}x\left(1+\frac{\sin^{2}x}{\cos^{2}}\right)=\tan^{2}x

\sin^{2}x\left(1\times\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}}\right)=\tan^{2}x

 

[tornzaer]

Alright that clears up a lot. So I can make a 1 into cos/cos or sin/sin. Makes things a lot easier. Thank you.

 

[unplebeian]

Tornaer:

there is an identity relating to sec^x and 1+tan^2x. Find it out and you'll get your answer in 2 lines or less.