Question

Match the following
Given $\sin A=\frac{2}{3}\text{and}\sin B=\frac{1}{4}$
$A$ and $B$ are acute angles.
$\begin{array}{cc}\left(1\right)\sin (A+B)& \left(p\right)\frac{2\sqrt{15}-\sqrt{5}}{2+5\sqrt{3}}\\ \left(2\right)\cos (A-B)& \left(q\right)\frac{55}{144}\\ \left(3\right)\tan (A-B)& \left(r\right)\frac{2\sqrt{15}+\sqrt{5}}{12}\\ \left(4\right)\sin (A+B)\sin (A-B)& \left(s\right)\frac{5\sqrt{3}+2}{12}\end{array}$

• A. 1-r, 2-p, 3-s, 4-q
• B. 1-s, 2-r, 3-q, 4-p
• C. 1-s, 2-r, 3-p, 4-q
• D. 1-r, 2-s, 3-p, 4-q

Answer 1

The correct option is D

1-r, 2-s, 3-p, 4-q

This is a direct application of $\sin (A+B),\sin (A-B)$, $\cos (A+B)$ etc., or trigonometric ratios of compound angles.

Before we calculate them, we need $\cos A,\cos B,\tan A$ and $\tan B$ to easily calculate the ratios asked. For that, we will construct proper triangles and find the ratios.

$\sin A=\frac{2}{3}\text{and}\sin B=\frac{1}{4}$

$\Rightarrow \cos A=\frac{\sqrt{5}}{3}\text{and}\cos B=\frac{\sqrt{15}}{4}$

$\Rightarrow \tan A=\frac{2}{\sqrt{5}}\text{and}\tan B=\frac{1}{\sqrt{15}}$

(1) $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$=\frac{2}{3}\times \frac{\sqrt{15}}{4}+\frac{\sqrt{5}}{3}\times \frac{1}{4}$

= $\frac{2\sqrt{15}+\sqrt{5}}{12}$

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

$=\frac{\sqrt{5}}{3}\times \frac{\sqrt{15}}{4}+\frac{2}{3}\times \frac{1}{4}$

= $\frac{5\sqrt{3}+2}{12}$

(3) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

= $\frac{\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{15}}}{(1+\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{15}})}$

$=\frac{2\sqrt{15}-\sqrt{5}}{\sqrt{5}\times \sqrt{15}+2}$

$=\frac{2\sqrt{15}-\sqrt{5}}{5\sqrt{3}+2}$

(4) $\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B$

$={\left(\frac{2}{3}\right)}^2-{\left(\frac{1}{4}\right)}^2$

$=\frac{4}{9}-\frac{1}{16}$

$=\frac{64-9}{9\times 16}$

$=\frac{55}{144}$