Question

limit of sin^2 theta/ tan^2 theta as theta approaches pi/2

Answer 1

Recall that \(\large \tan^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta}\). Thus, we note that \(\large \displaystyle\lim_{\theta\to\frac{\pi}{2}} \frac{\sin^2\theta}{\tan^2\theta} = \lim_{\theta\to\frac{\pi}{2}} \sin^2\theta\cdot \frac{\cos^2\theta}{\sin^2\theta}=\ldots\).

Can you take things from here? :-)

Answer 2

cancel sin^2 theta

Answer 3

the answer should be 1/2.. and i dont know how to get this..

Answer 4

So what answer are you getting after you have cancelled out the squared sines?

Answer 5

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