Question

Match the following
Given $sinA=\frac{2}{3}andsinB=\frac{1}{4}$

$\begin{array}{cc}\left(1\right)sin(A+B)& \left(p\right)\frac{2\sqrt{15}-\sqrt{5}}{2+5\sqrt{3}}\\ \left(2\right)Cos(A-B)& \left(q\right)\frac{55}{144}\\ \left(3\right)Tan(A-B)& \left(r\right)\frac{2\sqrt{15}+\sqrt{5}}{12}\\ \left(4\right)Sin(A+B)sin(A-b)& \left(s\right)\frac{5\sqrt{3}+2}{12}\end{array}$

• 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 cosA, cosB, tanA and tanB to easily calculate the ratios asked. For that, we will construct proper triangles and find the ratios.

$SinA=\frac{2}{3}$ $SinB=\frac{1}{4}$

$cosA=\frac{\sqrt{5}}{3}$ $cosA=\frac{\sqrt{15}}{4}$

$tanA=\frac{2}{\sqrt{5}}$ $tanB=\frac{1}{\sqrt{15}}$

(1) Sin (A+B) = sinA cosB + cosA sinB

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

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

(2) Cos(A-B) = cosA cosB + sinA sinB

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

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

(3) tan(A - B) = $fractanA-tanB1+tanAtanB$

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

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

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

(4) Sin(A+B) sin(A-B) = $si{n}^{2}A-si{n}^{2}B$

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

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

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

= $\frac{55}{144}$