Q.6. Let tan A, tan B, where , be the roots of the quadratic equation . Then is equal to :

Q: What is the correct answer for JEE Main Mathematics Q6?

The correct answer is C. Sol. $x^2 - 2x - 5 = 0$From Vieta's formulas:$\tan A + \tan B = 2$$\tan A \tan B = -5$Now, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$We know that $\cos(A+B) = \frac{1}{\sec(A+B)} = \frac{1}{\pm\sqrt{1+\tan^2(A+B)}}$.Since $A, B \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, $A+B \in (-\pi, \pi)$.If $\tan(A+B) = 1/3 > 0$, then $A+B$

Q: Q.6. Let tan A, tan B, where , be the roots of the quadratic equation . Then is equal to :

The correct answer is C. Sol. $x^2 - 2x - 5 = 0$From Vieta's formulas:$\tan A + \tan B = 2$$\tan A \tan B = -5$Now, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$We know that $\cos(A+B) = \frac{1}{\sec(A+B)} = \frac{1}{\pm\sqrt{1+\tan^2(A+B)}}$.Since $A, B \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, $A+B \in (-\pi, \pi)$.If $\tan(A+B) = 1/3 > 0$, then $A+B$

Q: What are the options for this JEE Main Mathematics question?

A. $10+\sqrt{10}$ · B. $10-2\sqrt{10}$ · C. $10-3\sqrt{10}$ · D. $10-4\sqrt{10}$

Question

Q.6. Let tan A, tan B, where , be the roots of the quadratic equation . Then  is equal to :

FirstInTest · Accepted Answer

Sol. $x^2 - 2x - 5 = 0$From Vieta's formulas:$	an A + 	an B = 2$$	an A 	an B = -5$Now, $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$We know that $\cos(A+B) = \frac{1}{\sec(A+B)} = \frac{1}{\pm\sqrt{1+	an^2(A+B)}}$.Since $A, B \in \left( -\frac{\pi}{2}, \frac{\pi}{2} ight)$, $A+B \in (-\pi, \pi)$.If $	an(A+B) = 1/3 > 0$, then $A+B$ is in the first or third quadrant.Since $A, B \in \left( -\frac{\pi}{2}, \frac{\pi}{2} ight)$, $A+B \in (-\pi, \pi)$. If $A+B$ is in the third quadrant, $\cos(A+B)$ would be negative. If $A+B$ is in the first quadrant, $\cos(A+B)$ would be positive.The problem doesn't specify the quadrant for $A+B$. However, the options are positive, suggesting we should take the positive root for $\cos(A+B)$.$\cos(A+B) = \frac{1}{\sqrt{1 + (1/3)^2}} = \frac{1}{\sqrt{1 + 1/9}} = \frac{1}{\sqrt{10/9}} = \frac{3}{\sqrt{10}}$We need to find $20\sin^2\left( \frac{A+B}{2} ight)$.Using the half-angle identity: $\sin^2	heta = \frac{1 - \cos(2	heta)}{2}$.Let $	heta = \frac{A+B}{2}$, so $2	heta = A+B$.$20\sin^2\left( \frac{A+B}{2} ight) = 20 \left( \frac{1 - \cos(A+B)}{2} ight)$$= 10 (1 - \cos(A+B))$$= 10 \left( 1 - \frac{3}{\sqrt{10}} ight)$$= 10 - \frac{30}{\sqrt{10}} = 10 - 3\sqrt{10}$

Answer

$10+\sqrt{10}$

Answer

$10-2\sqrt{10}$

Answer

$10-3\sqrt{10}$

Answer

$10-4\sqrt{10}$

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