trigonometric identity of sin squared in terms of tan squared.
Why is $\sin^2(x)=\frac{\tan^2(x)}{1+\tan^2(x)}$? And why is $\sin^2(x)=\frac{1}{\cot^2(x)}$?
I've tried starting from $\tan^2(x)=\frac{\sin^2(x)}{1-\sin^2(x)}$ but that wasn't really working out for me.
$\endgroup$ 43 Answers
$\begingroup$For example:
$$\frac{\tan^2x}{1+\tan^2x}=\frac{\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}=\frac{\sin^2x}{\cos^2x+\sin^2x}=\sin^2x$$
Now you try the other ones.
$\endgroup$ $\begingroup$$$\sin^2x=\frac{1}{\csc^2x}=\frac{\frac{1}{\csc^2x}}{1}\cdot\frac{\sec^2 x}{\sec^2 x}=\frac{(\frac{\sec x}{\csc x})^2}{\sec^2 x}=\frac{\tan^2 x}{\tan^2+1}$$
Now all that's left to prove is that $\tan x = \dfrac{\sec x}{\csc x}$ $$\tan x = \frac{\sin x}{\cos x}=\frac{\frac{1}{\csc x}}{\frac{1}{\sec x}}=\frac{\frac{1}{\csc x}}{\frac{1}{\sec x}}\cdot\frac{\frac{\sec x\csc x}{1}}{\frac{\sec x\csc x}{1}}=\frac{\sec x}{\csc x}$$
$\endgroup$ $\begingroup$Using the identity $\tan^2x + 1 = \sec^2x$ yields
\begin{align*} \frac{\tan^2x}{1 + \tan^2x} & = \frac{\tan^2x}{\sec^2x}\\ & = \tan^2x\cos^2x\\ & = \frac{\sin^2x}{\cos^2x} \cdot \cos^2x\\ & = \sin^2x \end{align*}
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