First slide

Trigonometric Identities

Question

If $x = a \cos^{3} θ \sin^{2} θ, y = a \sin^{3} θ \cos^{2} θ$ and $\frac{{(x^{2} + y^{2})}^{p}}{(x y)^{q}} (p, q, \in N)$ is independent of $θ$ , then

Moderate

A

$4 p = 5 q$
B

$4 q = 5 p$
C

$p + q = 9$
D

$p q = 20$

Solution

$\begin{array}{l} \frac{{(x^{2} + y^{2})}^{p}}{(x y)^{q}} = \frac{{[a \sin^{2} θ \cos^{2} θ]}^{2 p}}{{[a^{2} \sin^{5} θ \cos^{5} θ]}^{q}} \\ = \frac{a^{2 p} (\sin θ \cos θ)^{4 p}}{a^{2 q} (\sin θ \cos θ)^{5 q}} \end{array}$
which is independent of $θ$ if $4 p = 5 q$ .

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