Q102 / 180 · COMEDK 2022 · Mathematics
If $\sin A+\sin B=a$ and $\cos A+\cos B=b$, then $\cos (A+B)$ equals?
- A$\dfrac{a^2+b^2}{b^2-a^2}$
- B$\dfrac{2 a b}{a^2+b^2}$
- Correct answerC$\dfrac{b^2-a^2}{a^2+b^2}$
- D$\dfrac{a^2-b^2}{a^2+b^2}$
Solution
Given $\sin A + \sin B = a$ and $\cos A + \cos B = b$.
Squaring both equations:
$(\sin A + \sin B)^2 = a^2 \Rightarrow \sin^2 A + \sin^2 B + 2 \sin A \sin B = a^2$
$(\cos A + \cos B)^2 = b^2 \Rightarrow \cos^2 A + \cos^2 B + 2 \cos A \cos B = b^2$
Adding the two squared equations:
$(\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\cos A \cos B + \sin A \sin B) = a^2 + b^2$
$1 + 1 + 2 \cos(A - B) = a^2 + b^2$
$2 \cos(A - B) = a^2 + b^2 - 2$
Subtracting the squared equations ($b^2 - a^2$):
$b^2 - a^2 = (\cos^2 A + \cos^2 B + 2 \cos A \cos B) - (\sin^2 A + \sin^2 B + 2 \sin A \sin B)$
$b^2 - a^2 = (\cos^2 A - \sin^2 A) + (\cos^2 B - \sin^2 B) + 2(\cos A \cos B - \sin A \sin B)$
$b^2 - a^2 = \cos 2A + \cos 2B + 2 \cos(A + B)$
$b^2 - a^2 = 2 \cos(A + B) \cos(A - B) + 2 \cos(A + B)$
$b^2 - a^2 = 2 \cos(A + B) [\cos(A - B) + 1]$
From $2 \cos(A - B) = a^2 + b^2 - 2$, we have $\cos(A - B) = \dfrac{a^2 + b^2 - 2}{2}$.
Substituting this into the expression for $b^2 - a^2$:
$b^2 - a^2 = 2 \cos(A + B) [\dfrac{a^2 + b^2 - 2}{2} + 1]$
$b^2 - a^2 = 2 \cos(A + B) [\dfrac{a^2 + b^2}{2}]$
$b^2 - a^2 = \cos(A + B) (a^2 + b^2)$
$\cos(A + B) = \dfrac{b^2 - a^2}{a^2 + b^2}$
Answer: $\dfrac{b^2-a^2}{a^2+b^2}$
Squaring both equations:
$(\sin A + \sin B)^2 = a^2 \Rightarrow \sin^2 A + \sin^2 B + 2 \sin A \sin B = a^2$
$(\cos A + \cos B)^2 = b^2 \Rightarrow \cos^2 A + \cos^2 B + 2 \cos A \cos B = b^2$
Adding the two squared equations:
$(\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\cos A \cos B + \sin A \sin B) = a^2 + b^2$
$1 + 1 + 2 \cos(A - B) = a^2 + b^2$
$2 \cos(A - B) = a^2 + b^2 - 2$
Subtracting the squared equations ($b^2 - a^2$):
$b^2 - a^2 = (\cos^2 A + \cos^2 B + 2 \cos A \cos B) - (\sin^2 A + \sin^2 B + 2 \sin A \sin B)$
$b^2 - a^2 = (\cos^2 A - \sin^2 A) + (\cos^2 B - \sin^2 B) + 2(\cos A \cos B - \sin A \sin B)$
$b^2 - a^2 = \cos 2A + \cos 2B + 2 \cos(A + B)$
$b^2 - a^2 = 2 \cos(A + B) \cos(A - B) + 2 \cos(A + B)$
$b^2 - a^2 = 2 \cos(A + B) [\cos(A - B) + 1]$
From $2 \cos(A - B) = a^2 + b^2 - 2$, we have $\cos(A - B) = \dfrac{a^2 + b^2 - 2}{2}$.
Substituting this into the expression for $b^2 - a^2$:
$b^2 - a^2 = 2 \cos(A + B) [\dfrac{a^2 + b^2 - 2}{2} + 1]$
$b^2 - a^2 = 2 \cos(A + B) [\dfrac{a^2 + b^2}{2}]$
$b^2 - a^2 = \cos(A + B) (a^2 + b^2)$
$\cos(A + B) = \dfrac{b^2 - a^2}{a^2 + b^2}$
Answer: $\dfrac{b^2-a^2}{a^2+b^2}$
Hence, the correct answer is option: 'C'.