Question

Prove that \[\sec({1\over2}\pi-x) = cosec ~x\]

Answer 1

\[\frac{1}{\cos(\pi/2-x)} = \csc x\]\[\frac{1}{\cos(\pi/2) \cos(x) + \sin(\pi/2)\sin(x)} = \csc x\]
cos(pi/2)=0 and sin(pi/2)=1 so\[\frac{1}{\sin(x)} =\csc x\]\[\csc x = \csc x\]

Answer 2

Btw, that 2nd step used the cosine subtraction identity: \[\cos(A-B) = \cos A \cos B + \sin A \sin B\]

Answer 3

you can also use the complementary angle property

sin(x) = cos(pi/2 - x) works for reciprocals as well

Answer 4

Oh, I had never heard of the complementary angle property. Thanks :)

Answer 5

Thanks. I understand Alexray's more though. But I can always try different methods.

Answer 6

See the above diagram.
\[\sec (\pi/2-x)=OB/OC\]
\[\csc x=OB/AB\]
Now, you can see that OABC is a rectangle, so AB=OC.
So, \[\sec (\pi/2-x)=\csc x\]
I know that this has already been done, but I just wanted to put in my way.