First slide

Trigonometric Identities

Question

If $x = h + p \sec α, y = k + q \cos e c α$ then ${(\frac{p}{x - h})}^{2} + {(\frac{q}{y - k})}^{2} =$

Easy

A

1
B

-1
C

0
D

$\frac{1}{2}$

Solution

$W e h a v e x = h + p s e c α a n d y = k + q \cos e c α . \Rightarrow \frac{x - h}{p} = s e c α, \frac{y - k}{q} = \cos e c α \Rightarrow \frac{p}{x - h} = \cos α, \frac{q}{y - k} = \sin α \Rightarrow {(\frac{p}{x - h})}^{2} + {(\frac{q}{y - k})}^{2} = \cos^{2} α + \sin^{2} α = 1$

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