1(ii) Explain why tan x > sin x for 0<x<90

Yes

Yes

Sorry, didn't see this as I wasn't quoted.

So because the interval is 0 < x < 90, you know that the RHS is always going to be positive as both sin(x) and tan(x) are both positive for this range, so sin²(x) and tan²(x) will be too.

Factorise the LHS and you'll probably find your answer

I'm not sure how to factorise the LHS.

Are you familiar with the difference of two squares?

Ohhhh I get the factorisation now

Great so what do you spot?

What am I meant to spot?

We've established that the RHS is always positive for values of x in this range. Because of the identity, the LHS must also be positive. Look at the result of your factorisation, keeping this in mind, and things should become reasonably clear

Thank you!

...

We know that the RHS is $> 0$ for $0< x < 90$ because it's a whole square term (anything squared is always +ve)

Just a pedantic quibble on language but anything squared is always non-negative, not necessarily positive; it can be zero, and the fact that it's not in this case is important to the argument.

Just a pedantic quibble on language but anything squared is always non-negative, not necessarily positive; it can be zero, and the fact that it's not in this case is important to the argument.

For more pedantry: any real number squared is always non-negative; i^2 = -1.