First slide

Multiple and sub- multiple Angles

Question

If $\tan θ = \frac{p}{q}$ and $θ = 3 ϕ (0 < θ < \frac{π}{2})$ , then $\frac{p}{\sin ϕ} - \frac{q}{\sin ϕ} =$

Moderate

A

$2 \sqrt{(p^{2} - q^{2})}$
B

$2 \sqrt{(p^{2} + q^{2})}$
C

$\sqrt{(p^{2} + q^{2})}$
D

$3 \sqrt{(p^{2} + q^{2})}$

Solution

From given relation $\frac{\sin θ}{p} = \frac{\cos θ}{q} = \sqrt{\frac{\sin^{2} θ + \cos^{2} θ}{p^{2} + q^{2}}} = \frac{1}{\sqrt{p^{2} + q^{2}}}$
Now putting for p and q in the L.H.S. we have
$L . H . S . = \sqrt{(p^{2} + q^{2})} [\frac{\sin θ}{\sin ϕ} - \frac{\cos θ}{\cos ϕ}]$
$= \sqrt{(p^{2} + q^{2})} \frac{\sin (θ - ϕ)}{\sin ϕ \cos ϕ}$
$= \sqrt{(p^{2} + q^{2})} .2 \frac{\sin (3 ϕ - ϕ)}{\sin 2 ϕ} = 2 \sqrt{p^{2} - q^{2}} ∵ θ = 3 ϕ$

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