If tan A and tan B are the roots of the quadr | vedclass

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(B) Given that $	an A$ and $	an B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$.From the properties of quadratic equations,we have

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$-25$

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(B) Given that $	an A$ and $	an B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$.From the properties of quadratic equations,we have $	an A + 	an B = \frac{10}{3}$ and $	an A 	an B = -\frac{25}{3}$.We know that $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{10/3}{1 - (-25/3)} = \frac{10/3}{28/3} = \frac{10}{28} = \frac{5}{14}$.Now,the expression is $E = 3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$.Dividing the entire expression by $\cos^2 (A + B)$,we get $E = \cos^2 (A + B) [3 	an^2 (A + B) - 10 	an (A + B) - 25]$.Since $	an (A + B) = \frac{5}{14}$,the term inside the bracket is $3(\frac{5}{14})^2 - 10(\frac{5}{14}) - 25 = 3(\frac{25}{196}) - \frac{50}{14} - 25 = \frac{75}{196} - \frac{700}{196} - \frac{4900}{196} = \frac{-5525}{196}$.Also,$\cos^2 (A + B) = \frac{1}{1 + 	an^2 (A + B)} = \frac{1}{1 + (5/14)^2} = \frac{1}{1 + 25/196} = \frac{196}{221}$.Therefore,$E = \frac{196}{221} 	imes \frac{-5525}{196} = -\frac{5525}{221} = -25$.

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$,then the value of $3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$ is

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