Trigonometric Identity: Tan^2-Sin^2 = Sin^2 Cos^2

[816318]

\tan\left({^2}\right)-\sin\left({^2}\right)=\tan\left({^2}\right) \sin\left({^2}\right)
i keep on getting \sin\left({^2}\right)-\sin\left({^2}\right) \cos\left({^2}\right)=\sin\left({^2}\right) \sin\left({^2}\right)
\cos\left({^2}\right) \cos\left({^2}\right)

 

[SuperSonic4]

[math]\tan(2\theta)-\sin(2\theta)=\tan(2\theta)\sin(2\theta)[/math]

i keep on getting [math]\dfrac{\sin(2\theta)-\sin(2\theta) \cos(2\theta)}{\cos(2\theta)}=\dfrac{\sin(2\theta)\sin(2\theta)}{\cos(2\theta)}[/math]

Do you mean

[math]\tan(2\theta)-\sin(2\theta)=\tan(2\theta)\sin(2\theta)[/math] OR [math]\tan^2(\theta)-\sin^2(\theta)=\tan^2(\theta)\sin^2(\theta)[/math]

It would also be better if you used a more descriptive title and explain how you go to your outcome and also what the question is askingedit: I've used the site's LaTeX feature to make it easier to read.

 

[soroban]

$$\tan^2\theta -\sin^2\theta \;=\; \tan^2\theta \sin^2\theta$$

$$\begin{array}{ccc}<br /> \tan^2\theta - \sin^2\theta &=& \dfrac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta \\ \\<br /> & = & \sin^2\theta\left(\dfrac{1}{\cos^2\theta} - 1\right) \\ \\<br /> <br /> & = & \sin^2\theta\left(\dfrac{1-\cos^2\theta}{\cos^2\theta}\right) \\ \\<br /> <br /> & = & \sin^2\theta \left(\dfrac{\sin^2\theta}{\cos^2\theta}\right) \\ \\<br /> <br /> & = & \left(\dfrac{\sin^2\theta}{\cos^2\theta}\right)\sin^2\theta \\ \\<br /> <br /> & = & \tan^2\theta\sin^2\theta\end{array}$$