Question

If x= sin theta then x/√1-x^2 is​

Answer 1

Well,

Given that $x=\sin(\theta)$,

We can rewrite the equation like,

$\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}$

Now use, $\cos(\theta)^2+\sin(\theta)^2=1$ which implies that $1-\sin(\theta)^2=\cos(\theta)^2$

That means that,

$\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}\Longleftrightarrow\dfrac{\sin(\theta)}{\sqrt{\cos(\theta)^2}}$

By def $\sqrt{x^2}=x$ therefore $\sqrt{\cos(\theta)^2}=\cos(\theta)$

So the fraction now looks like,

$\dfrac{\sin(\theta)}{\cos(\theta)}$

Which is equal to the identity,

$\boxed{\tan(\theta)}=\dfrac{\sin(\theta)}{\cos(\theta)}$

Hope this helps.

r3t40