Question

Help please will medal

Answer 1

@iTired, Can you help with this?

Answer 2

@MyNameIsNemo

Answer 3

Idk that Class XD

Answer 4

try to use this identity:
\[\huge \cos \left( {\frac{\pi }{2} - x} \right) = \sin x\]

Answer 5

See I have a problem, I don't know how to work this

Answer 6

If I apply my identity above, I get:
\[\Large \frac{{\cos {{\left( {\frac{\pi }{2} - x} \right)}^2}}}{{\sqrt {1 - {{\left( {\sin x} \right)}^2}} }} = \frac{{{{\left( {\sin x} \right)}^2}}}{{\cos x}} = ...?\]
since:
\[\Large \sqrt {1 - {{\left( {\sin x} \right)}^2}} = \cos x\]

Answer 7

So this probably would have been smart to give in the first place but how would I get it to this?

Answer 8

you have to apply this identity:
\[\Large \frac{{\sin x}}{{\cos x}} = \tan x\]
please keep in mind that we can write this:
\[\Large \frac{{\cos {{\left( {\frac{\pi }{2} - x} \right)}^2}}}{{\sqrt {1 - {{\left( {\sin x} \right)}^2}} }} = \frac{{{{\left( {\sin x} \right)}^2}}}{{\cos x}} = \frac{{\sin x}}{{\cos x}} \cdot \sin x = ...?\]