If A and B (A B) are acute angles,sin (A-B) | vedclass

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Question

(A) Given: $\sin B = \frac{5}{13}$. Since $B$ is an acute angle,$\cos B = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$. Thus,$	an B = \frac{5}{12}$.G

Vedclass · Accepted Answer

$\frac{714025}{342732}$

Vedclass · Answer

(A) Given: $\sin B = \frac{5}{13}$. Since $B$ is an acute angle,$\cos B = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$. Thus,$	an B = \frac{5}{12}$.Given: $\sin (A-B) = \frac{16}{65}$. Since $A$ and $B$ are acute,$\cos (A-B) = \sqrt{1 - (\frac{16}{65})^2} = \sqrt{\frac{4225-256}{4225}} = \sqrt{\frac{3969}{4225}} = \frac{63}{65}$.Thus,$	an (A-B) = \frac{16}{63}$.Using the formula $	an (A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have:$\frac{	an A - \frac{5}{12}}{1 + 	an A \cdot \frac{5}{12}} = \frac{16}{63}$$\frac{12 	an A - 5}{12 + 5 	an A} = \frac{16}{63}$$63(12 	an A - 5) = 16(12 + 5 	an A)$$756 	an A - 315 = 192 + 80 	an A$$676 	an A = 507$$	an A = \frac{507}{676} = \frac{3}{4}$.Then $\cot A = \frac{1}{	an A} = \frac{4}{3}$.Finally,$	an A + \cot A = \frac{3}{4} + \frac{4}{3} = \frac{9+16}{12} = \frac{25}{12}$.Wait,re-evaluating the calculation: $	an A = \frac{507}{676} = 0.75 = \frac{3}{4}$. The provided option $\frac{714025}{342732}$ is $\frac{507^2 + 676^2}{507 	imes 676} = \frac{257049 + 456976}{342732} = \frac{714025}{342732}$. This matches option $A$.

If $A$ and $B$ $(A > B)$ are acute angles,$\sin (A-B)=\frac{16}{65}$ and $\sin B=\frac{5}{13}$,then $\tan A+\cot A=$

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