## Trigonometry (11th Edition) Clone

$$\tan^2\alpha\sin^2\alpha=\tan^2\alpha+\cos^2\alpha-1$$ We examine the right side first, which eventually shows that the expression is an identity.
$$\tan^2\alpha\sin^2\alpha=\tan^2\alpha+\cos^2\alpha-1$$ We examine the right side first. $$A=\tan^2\alpha+\cos^2\alpha-1$$ - We can rewrite $1$ as follows: $$1=\sin^2\alpha+\cos^2\alpha$$ That makes $A$ into $$A=\tan^2\alpha+\cos^2\alpha-(\sin^2\alpha+\cos^2\alpha)$$ $$A=\tan^2\alpha+\cos^2\alpha-\sin^2\alpha-\cos^2\alpha$$ $$A=\tan^2\alpha-\sin^2\alpha$$ - Now, we would rewrite $\tan^2\alpha$: $$\tan^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}$$ Then $A$ is $$A=\frac{\sin^2\alpha}{\cos^2\alpha}-\sin^2\alpha$$ $$A=\frac{\sin^2\alpha-\sin^2\alpha\cos^2\alpha}{\cos^2\alpha}$$ $$A=\frac{\sin^2\alpha(1-\cos^2\alpha)}{\cos^2\alpha}$$ - Then we rewrite $$1-\cos^2\alpha=\sin^2\alpha$$ $$A=\frac{\sin^2\alpha\sin^2\alpha}{\cos^2\alpha}$$ $$A=\Big(\frac{\sin\alpha}{\cos\alpha}\Big)^2\sin^2\alpha$$ $$A=\tan^2\alpha\sin^2\alpha$$ (since $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$) 2 sides are proved to be equal and the expression is an identity.