Question

EXPLAIN THE SECOND PART.......

Answer 1

$\tan^2\theta-\sin^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta$
$=\sin^2\theta\left(\dfrac1{\cos^2\theta}-1\right)$
$=\sin^2\theta(\sec^2\theta-1)$
$=\sin^2\theta\tan^2\theta$

Now if $0^\circ<\theta<90^\circ$, then both $\sin\theta>0$ and $\tan\theta>0$. Adding two positive numbers gives another positive number, so $\tan\theta+\sin\theta>0$. Why this is useful will be apparent shortly.

Back to the identity:

$\tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta$

Factorizing the left hand side, we have

$(\tan\theta-\sin\theta)(\tan\theta+\sin\theta)=\tan^2\theta\sin^2\theta$

Now, any number squared will be positive, which means the right hand side is necessarily greater than 0.

We showed earlier that $\tan\theta+\sin\theta>0$. So we understand that we have

$\underbrace{(\tan\theta-\sin\theta)}_?\underbrace{(\tan\theta+\sin\theta)}_+=\underbrace{\tan^2\theta\sin^2\theta}_+$

The only way to multiply a number by a positive number to get yet another positive number is to have the first number also be positive, which means

$\tan\theta-\sin\theta>0$

and from this it follows that

$\tan\theta>\sin\theta$

in the provided region.