Question

The tangents of two acute angles are 3 and 2. The sine of twice their difference is

• A. 7/24
• B. 7/48
• C. 7/50
• D. 7/25

Answer 1

The correct option is D 7/25

$tanA=3\Rightarrow A=ta{n}^{-1}3$

$tanB=2\Rightarrow B=ta{n}^{-1}2$

Hence

$A-B$

$=ta{n}^{-1}\left(3\right)-ta{n}^{-1}\left(2\right)$

$=ta{n}^{-1}\left(\frac{3-2}{1+6}\right)$
$=ta{n}^{-1}\left(\frac{1}{7}\right)$
$=si{n}^{-1}\left(\frac{1}{5\sqrt{2}}\right)$

Hence

$sin(A-B)=\frac{1}{5\sqrt{2}}$

And

$cos(A-B)=\frac{7}{5\sqrt{2}}$

Hence

$sin\left(2\right(A-B\left)\right)$

$=2sin(A-B)cos(A-B)$

$=\frac{14}{50}$
$=\frac{7}{25}$.