How do you use sum and difference identities to simplify $csc (\frac{π}{2} + x) - sec (π - x)$ $csc(pi/2 +x) - sec(pi - x)$ ?

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How do you use sum and difference identities to simplify $csc (\frac{π}{2} + x) - sec (π - x)$ $csc(pi/2 +x) - sec(pi - x)$ ?

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Answer 1 · 2018-03-14T16:31:40

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Explanation:

$csc (θ) = \frac{1}{sin (θ)}, sec (θ) = \frac{1}{cos (θ)}$ $csc(theta) = 1/sin(theta) , sec(theta) = 1/cos(theta)$
$\frac{1}{sin (\frac{π}{2} + x)} - \frac{1}{cos (π - x)}$ $1/(sin(pi/2 + x)) - 1/(cos(pi - x))$

$sin (a + b) = sin a \cdot cos b + cos a \cdot sin b$ $sin(a+b) = sina*cosb + cosa * sin b$
$cos (a - b) = cos a \cdot cos b + sin a \cdot sin b$ $cos(a - b) = cosa*cosb + sina * sinb$

$\frac{1}{sin (\frac{π}{2}) \cdot cos x + cos (\frac{π}{2}) \cdot sin x} - \frac{1}{cos π \cdot cos x + sin π \cdot sin x}$ $1/(sin(pi/2) *cosx + cos(pi/2)*sinx) - 1/(cospi*cosx + sinpi*sinx)$
Using either calculator or unit circle,
$cos (\frac{π}{2}) = 0, sin (π) = 0, cos (π) = 1, sin (\frac{π}{2}) = 1$ $cos(pi/2) = 0, sin(pi) = 0, cos(pi) = 1, sin(pi/2) = 1$

$\frac{1}{cos x} - \frac{1}{cos x} = 0$ $1/(cosx) - 1/(cosx) = 0$

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How do you use sum and difference identities to simplify $csc (\frac{π}{2} + x) - sec (π - x)$ $csc(pi/2 +x) - sec(pi - x)$ ?

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How do you use sum and difference identities to simplify $csc (\frac{π}{2} + x) - sec (π - x)$ $csc(pi/2 +x) - sec(pi - x)$ ?