Question

if A and B are acute angle satisfying 3sin^2A + 2 sin^2 B=1 and 3 sin 2A = 2 Sin 2B then cos(A+2B)=?​

Answer 1

Step-by-step explanation:

answer is the above image

Answer 2

Let $0 < A,B < \frac{n}{2}$ satisfying the equation

$3sin^2A+2sin^2B=1$ & $3sin2A-2sin2B=0$ then

$A+2B$ is equal to

$3sin^2 A +2sin^2 B =1$

$3sin^2 A=1-2sin^2B$

$B sinA=\frac{cos2B}{Sin A}$ .....(1)

$3sin2A-2sin2B=0$

$3[2sinA.cosA]-2sin2B=0$

$3sinA=\frac{sin2B}{cosA}$ .....(2)

$\frac{sin2B}{cosA}=\frac{cos2B}{sinA}$

$\frac{sin2B}{cos2B}=\frac{cosA}{sinA}$    $\therefore \frac{1}{tanA}$  $tan2B.tanA=1$

$tan(A+2B)=\frac{tanA+tan 2B}{1-tanA.tan2B}$

$tan(A+2B)=\frac{tanA+tan 2B}{0}$

$A+2B=\frac{r}{2}$

#SPJ2