Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to prove that: $$\tan^2\theta \sin^2\theta = \tan^2\theta - \sin^2 \theta$$ Here is what I have tried $$\tan^2\theta \sin^2\theta$$ $$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$ $$=\frac{\sin^4\theta}{\cos^2\theta}$$ Not much of an attempt, but now I am stuck. What should I do next? Thanks in advance for your answers ;)

share|cite|improve this question
$\sin^4 \theta = \sin^2 \theta \sin^2 \theta = (1 - \cos^2\theta)\sin^2\theta = \sin^2\theta - \sin^2\theta \cos^2\theta$. – Tunococ Mar 14 '14 at 0:38
up vote 2 down vote accepted

$$\tan^2\theta \sin^2\theta$$ $$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$ $$=\frac{\sin^4\theta}{\cos^2\theta}$$ $$=\frac{\sin^2\theta\sin^2\theta}{\cos^2\theta}$$ $$=\frac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}$$ $$=\frac{\sin^2\theta-\sin^2\theta\cos^2\theta}{\cos^2\theta}$$ $$=\frac{\sin^2\theta}{\cos^2\theta}-\frac{\sin^2\theta\cos^2\theta}{\cos^2\theta}$$ $$=\tan^2\theta-\sin^2\theta$$ $$\displaystyle \boxed{\therefore \tan^2\theta \sin^2\theta=\tan^2\theta - \sin^2\theta}$$

share|cite|improve this answer

$$\tan^2\theta \sin^2\theta+\sin^2 \theta=\sin^2\theta(\tan^2\theta +1) = \sin^2\theta\cdot\sec^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$

Alternatively, $$\frac1{ \sin^2\theta}-\frac1{\tan^2\theta}=\csc^2\theta-\cot^2\theta=1$$

share|cite|improve this answer
@user135339, how about this? – lab bhattacharjee Mar 14 '14 at 3:18


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.