If A and B are positive acute angle satisfying the equation 3cos2A+2cos2B=4 and sinB 3sinA =cosA 2cosB , then A+2B is

sinB 3sinA =cosA 2cosB ⇒cosA 3sinA =cosA 2cosB .cosA sinB ⇒tanA=3cos2A sin2B =cos2B sin2B .3cos2A cos2B =3cos2A tan2B(2cos2B−1) =3cos2A tan2B(4−3cos2A−1) (∵2cos2B=4−3cos2A )=tan2B.tan2A⇒tanA.tan2B=1⇒sinAsin2B=cosAcos2B ⇒cos(A+2B) =0⇒A+2B=2π

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If A and B are positive acute angle satisfying the equation 3

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If $A$ and $B$ are positive acute angle satisfying the equation $3 cos^{2} A + 2 cos^{2} B = 4$ and $\frac{3 sin A}{sin B} = \frac{2 cos B}{cos A},$ then $A + 2 B$ is

\frac{π}{2}

\frac{π}{3}

\frac{π}{4}

\frac{π}{6}

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A

and

B

are positive acute angle satisfying the equation

3 cos^{2} A + 2 cos^{2} B = 4

and

\frac{3 sin A}{sin B} = \frac{2 cos B}{cos A},

then

A + 2 B

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Text solutionVerified

\frac{3 sin A}{sin B} = \frac{2 cos B}{cos A} \Rightarrow \frac{3 sin A}{cos A} = \frac{2 cos B}{cos A} . \frac{sin B}{cos A}

\Rightarrow tan A = \frac{sin 2 B}{3 cos ^{2} A} = \frac{sin 2 B}{cos 2 B} . \frac{cos 2 B}{3 cos ^{2} A} = \frac{tan 2 B ( 2 cos ^{2} B - 1 )}{3 cos ^{2} A}

= \frac{tan 2 B ( 4 - 3 cos ^{2} A - 1 )}{3 cos ^{2} A} (∵ 2 cos^{2} B = 4 - 3 cos^{2} A)

= tan 2 B . tan^{2} A

\Rightarrow tan A . tan 2 B = 1 \Rightarrow sin A sin 2 B = cos A cos 2 B

\Rightarrow cos (A + 2 B) = 0 \Rightarrow A + 2 B = \frac{π}{2}

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Question Text	If $A$ and $B$ are positive acute angle satisfying the equation $3 cos^{2} A + 2 cos^{2} B = 4$ and $\frac{3 sin A}{sin B} = \frac{2 cos B}{cos A},$ then $A + 2 B$ is
Answer Type	Text solution:1
Upvotes	150

If A and B are positive acute angle satisfying the equation 3cos2A+2cos2B=4 and sinB 3sinA ​=cosA 2cosB ​, then A+2B is

Text solutionVerified

Practice questions from similar books

If $A$ and $B$ are positive acute angle satisfying the equation $3 cos^{2} A + 2 cos^{2} B = 4$ and $\frac{3 sin A}{sin B} = \frac{2 cos B}{cos A},$ then $A + 2 B$ is