sin⁡A=12\sin A = \frac{1}{\sqrt{2}}, sin⁡B=13\sin B = \frac{1}{\sqrt{3}} হলে tan⁡(A+B)= an(A + B) = কত?

Question

sin⁡A=12\sin A = \frac{1}{\sqrt{2}}, sin⁡B=13\sin B = \frac{1}{\sqrt{3}} হলে tan⁡(A+B)=	an(A + B) = কত?

Accepted Answer

ঘ. 2+12−1\frac{\sqrt{2} + 1}{\sqrt{2} - 1}

ব্যাখ্যা: দেওয়া আছে, $\sin A = \frac{1}{\sqrt{2}}$ এবং $\sin B = \frac{1}{\sqrt{3}}$।
সূক্ষ্মকোণের ক্ষেত্রে:
$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}} \implies 	an A = \frac{\sin A}{\cos A} = 1$
$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} \implies 	an B = \frac{\sin B}{\cos B} = \frac{1/\sqrt{3}}{\sqrt{2}/\sqrt{3}} = \frac{1}{\sqrt{2}}$
এখন, $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{1 + \frac{1}{\sqrt{2}}}{1 - 1 \cdot \frac{1}{\sqrt{2}}} = \frac{\frac{\sqrt{2} + 1}{\sqrt{2}}}{\frac{\sqrt{2} - 1}{\sqrt{2}}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$।

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