જો frac{1}{2} sin^{-1}left(frac{3 sin 2theta} | vedclass

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(B) ધારો કે $y = \frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2	heta}{5+4 \cos 2	heta}ight)$.તેથી $2y = \sin^{-1}\left(\frac{3 \sin 2	heta}{5+4 \cos 2

Vedclass · Accepted Answer

$\frac{1}{3} 	an 	heta$

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(B) ધારો કે $y = \frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2	heta}{5+4 \cos 2	heta}ight)$.તેથી $2y = \sin^{-1}\left(\frac{3 \sin 2	heta}{5+4 \cos 2	heta}ight)$,એટલે કે $\sin 2y = \frac{3 \sin 2	heta}{5+4 \cos 2	heta}$.નિત્યસમ $\sin 2y = \frac{2 	an y}{1+	an^2 y}$ અને $\sin 2	heta = \frac{2 	an 	heta}{1+	an^2 	heta}$,$\cos 2	heta = \frac{1-	an^2 	heta}{1+	an^2 	heta}$ નો ઉપયોગ કરતા:$\frac{2 	an y}{1+	an^2 y} = \frac{3 \left(\frac{2 	an 	heta}{1+	an^2 	heta}ight)}{5+4 \left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight)} = \frac{6 	an 	heta}{5(1+	an^2 	heta) + 4(1-	an^2 	heta)} = \frac{6 	an 	heta}{9+	an^2 	heta}$.ધારો કે $t = 	an 	heta$ અને $x = 	an y$. તેથી $\frac{2x}{1+x^2} = \frac{6t}{9+t^2}$.ગુણાકાર કરતા: $2x(9+t^2) = 6t(1+x^2) \implies 18x + 2xt^2 = 6t + 6tx^2$.ગોઠવતા: $6tx^2 - (18+2t^2)x + 6t = 0 \implies 3tx^2 - (9+t^2)x + 3t = 0$.અવયવ પાડતા: $(3x-t)(tx-3) = 0$.આમ,$x = \frac{t}{3} = \frac{1}{3} 	an 	heta$.

જો $\frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2\theta}{5+4 \cos 2\theta}\right) = \tan^{-1} x$ હોય,તો $x =$

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