I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this.
$$\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$$
I would really appreciate some inputs!
Best Answer
$$\frac{\cos^2\theta + \frac{\sin^2 \theta}{\cos^2\theta}-1}{\sin^2\theta}=$$ $$= \frac{\cos^2 \theta +\frac{\sin^2\theta}{\cos^2 \theta}-\sin^2 \theta-\cos^2\theta}{\sin^2 \theta}$$ $$= \frac{\frac{\sin^2\theta-\cos^2\theta \cdot \sin^2\theta}{\cos^2\theta}}{\sin^2\theta}$$ $$=\frac{\sin^2\theta \cdot (1-\cos^2\theta)}{\sin^2\theta \cdot \cos^2\theta}$$ $$=\frac{\sin^2\theta \cdot \sin^2 \theta}{sin^2\theta \cdot \cos^2\theta}$$ $$=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$