Question

$\text{If cos } \alpha + \text{cos } \beta = \text{a} \text{, } \text{sin } \alpha + \text{sin } \beta = \text{b } \text{and } \alpha - \beta = 2 \, \theta \text{, } \text{then } \frac{\text{cos } 3 \theta }{\text{cos } \theta } is equal to$

• A. $a^{2}+b^{2}-2$
• B. $a^{2}+b^{2}-3$
• C. $3-a^{2}-b^{2}$
• D. $\frac{a^{2} + b^{2}}{4}$

Answer 1

Answer: (B)

$\frac{\cos 3 \theta}{\cos \theta}=4 \cos ^{2} \theta-3=2(1+\cos 2 \theta)-3$
$=2 \cos 2 \theta-1=2 \cos (\alpha-\beta)-1$
Now, $\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)+\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+2 \cos (\alpha-\beta)= a ^{2}+ b ^{2}$
$\Rightarrow 2 \cos (\alpha-\beta)= a ^{2}+ b ^{2}-2$
$\Rightarrow \quad \frac{\cos 3 \theta}{\cos \theta}= a ^{2}+ b ^{2}-3$