If frac{tan 3A}{tan A} = a,then frac{sin 3A}{ | vedclass

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(B) Given that $\frac{	an 3A}{	an A} = a$.Using the formula $	an 3A = \frac{3	an A - 	an^3 A}{1 - 3	an^2 A}$,we have:$\frac{3	an A - 	an^3 A}{

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$\frac{2a}{a-1}$

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(B) Given that $\frac{	an 3A}{	an A} = a$.Using the formula $	an 3A = \frac{3	an A - 	an^3 A}{1 - 3	an^2 A}$,we have:$\frac{3	an A - 	an^3 A}{	an A(1 - 3	an^2 A)} = a$$\Rightarrow \frac{3 - 	an^2 A}{1 - 3	an^2 A} = a$$\Rightarrow 3 - 	an^2 A = a - 3a	an^2 A$$\Rightarrow 	an^2 A(3a - 1) = a - 3$$\Rightarrow 	an^2 A = \frac{a - 3}{3a - 1}$Now,consider $\frac{\sin 3A}{\sin A} = \frac{3\sin A - 4\sin^3 A}{\sin A} = 3 - 4\sin^2 A$.Using $\sin^2 A = \frac{	an^2 A}{1 + 	an^2 A}$,we get:$\sin^2 A = \frac{\frac{a-3}{3a-1}}{1 + \frac{a-3}{3a-1}} = \frac{a-3}{3a-1+a-3} = \frac{a-3}{4a-4} = \frac{a-3}{4(a-1)}$.Substituting this into the expression:$\frac{\sin 3A}{\sin A} = 3 - 4\left(\frac{a-3}{4(a-1)}ight) = 3 - \frac{a-3}{a-1}$$= \frac{3(a-1) - (a-3)}{a-1} = \frac{3a - 3 - a + 3}{a-1} = \frac{2a}{a-1}$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

If $\frac{\tan 3A}{\tan A} = a$,then $\frac{\sin 3A}{\sin A}$ is equal to

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