Question

Prove csc(pi/2-x)=sec x.

a. csc(pi/2-x)=1/sin(pi/2)cosx+cos(pi/2)cosx=sec x
b. csc(pi/2-x)=1/sin(pi/2)sinx-cos(pi/2)cosx=sec x
c. csc(pi/2-x)=1/sin(pi/2)sinx+cos(pi/2)cosx=sec x
d. csc(pi/2-x)=1/sin(pi/2)cosx-cos(pi/2)sinx=sec x

Answer 1

preliminary discussion:
complementary angles are a pair of angles that add up to 90°
btw, π/2 radians = 90°

there are 3 pairs of trig functions that are co-functions:
sine vs cosine
tangent vs cotangent
secant vs cosecant

e.g. sin20° = cos 70° , tan 28° = cot 62° , sec 10.2 = csc 79.8°
try them on your calculator.

I guess we are to prove one of them:

csc(pi/2-x)=sec x

LS =1/(sin(π/2) - x)
= 1/((sin(π/2)cosx - cos(π/2)sinx)
= 1/( 1(cosx - 0(sinx))
= 1/cosx
= sec x
= RS

A 5 line proof.

I have no idea what your choices are supposed to be.

Answer 2

the co- means complement.
so, cosec(x) = sec(x-complement) = sec(pi/2-x)
and so forth

If you play around with a right triangle, where the two acute angles are complementary, you can see how it all works.

Answer 3

In short, it's D. csc(pi/2-x)=1/sin pi/2 cos x-cos pi/2 sin x=sec x
It's on brainly too, so check it if you don't believe me. You're welcome, have a nice day <3