Question

(csc(x)-sec(x))/(csc(x)sec(x))

Answer 1

cosx-sinx

Answer 2

how?

Answer 3

\[\displaystyle \frac{a - b}{ab} = \frac{\cancel{a}}{\cancel{a}b} - \frac{\cancel{b}}{a\cancel{b}} = \frac{1}{b} - \frac{1}{a}\]

Here, \( a = \csc {x}\) and \(b = \sec {x} \), so it is equal to \(\displaystyle \frac{1}{\sec {x}} - \frac{1}{\csc {x}} = \boxed {\cos {x} - \sin {x}}\)

Answer 4

Thank you!
What about ((1-cos(x)^2)+sin^2(x))/(cos(x)-1)

Answer 5

sorry i wrote that wrong. it should be:
((1-cos(x))^2)+sin^2(x))/(cos(x)-1)

Answer 6

\((1-\cos x)^2 = 1 - 2\cos x + \cos^2 x\) and the identity that \(\cos^2 x + \sin^2 x = 1\).

So your expression reduces to: \(\displaystyle \frac{1 + (\cos^2 x + \sin^2 x) - 2\cos x}{\cos x - 1} = \frac{2 - 2\cos x}{\cos x - 1} = \frac{-2\cancel{(\cos x - 1)}}{\cancel{\cos x - 1}} = -2\)

Answer 7

thank you!
what about this one:
Verify (sec(m)+1)/(tan(m))=csc(m)+cot(m)