If sin theta = frac{3}{5}, then the value of | vedclass

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(B) Given $\sin 	heta = \frac{3}{5}.$Using the identity $\cos^2 	heta + \sin^2 	heta = 1,$ we have $\cos 	heta = \sqrt{1 - \sin^2 	heta}.$$\cos \

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$\frac{31}{60}$

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(B) Given $\sin 	heta = \frac{3}{5}.$Using the identity $\cos^2 	heta + \sin^2 	heta = 1,$ we have $\cos 	heta = \sqrt{1 - \sin^2 	heta}.$$\cos 	heta = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.$Now,calculate the other trigonometric ratios:$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{3/5}{4/5} = \frac{3}{4}.$$\cot 	heta = \frac{1}{	an 	heta} = \frac{4}{3}.$$\operatorname{cosec} 	heta = \frac{1}{\sin 	heta} = \frac{5}{3}.$Substitute these values into the expression $\frac{	an 	heta + \cos 	heta}{\cot 	heta + \operatorname{cosec} 	heta}$:Numerator: $\frac{3}{4} + \frac{4}{5} = \frac{15 + 16}{20} = \frac{31}{20}.$Denominator: $\frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3.$Result: $\frac{31/20}{3} = \frac{31}{60}.$

If $\sin \theta = \frac{3}{5},$ then the value of $\frac{\tan \theta + \cos \theta}{\cot \theta + \operatorname{cosec} \theta}$ is equal to

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