1. ## Solve

2. 1st you divide both left and right by cos x which gives you:

(sin x/cos x) > (cos x/cos x)

and we recognize that the left term will now be tan x

tan x > 1
x > arc tan 1
x > 45 degree

3. Not true. Take $x = \pi = 180^{o} > 45^{o}$ yet we have that $\tan 180^{o} = 0 < 1$

Imagining the unit circle would help and then generalize the case.

4. Hello,

You must differentiate the case when cos(x) is < 0, in which case the inequality will change its direction.

And as tan is periodic, let's restrict this to [0;2pi[

5. given - \pi <x> \pi

6. hmm good point lemme see,

tan x > 1

whereby tangent x is positive (shown on right hand side by the digit 1) hence we should consider the 1st and 3rd quardrant to find its limits.

x = 45 degree (lower limit)
x = 225 degree (upper limit)

hence the range it can go is

45 < x < 225 (if we only consider what the question wants and by inputing values into the calculator and comparing for sin and cos)

if we were to consider tan x>1 then the limits will be:
45 < x < 90
and
180 < x < 225