Trigonometry & Inverse Trigonometry
Trig Ratios Functions Identities
nta_abhyas_2025
Grade None
Question:
If $\cos \alpha + \cos \beta = a$, $\sin \alpha + \sin \beta = b$ and $\alpha - \beta = 2\theta$, then $\tan \frac{\alpha}{\tan \frac{\alpha}{2}} = \frac{a^2 + b^2 - 3}{\text{}}$
a^2 + b^2 - 2
a^2 + b^2 - 3
3 - a^2 - b^2
a^2+b^2/4
Step-by-Step Solution
Key Concept: Apply the product-to-sum formulas and use the constraint $a^2 + b^2 = 3$ to find the trigonometric ratio
Given $a^2 + b^2 = 3$, we need to find $\frac{\cos\alpha}{\cos\beta}$. We observe that $\cos\alpha / \cos\beta = 4\cos^2\theta - 3 = 2(1 + \cos 2\theta) - 3 = 2\cos(2\alpha - \beta) - 1$. Using the identity that $\cos^2\alpha + \sin^2\alpha = 1$ and $\cos^2\beta + \sin^2\beta = 1$, along with $2\cos(\alpha - \beta) - 1 = 2\cos(\alpha - \beta) - 1$. Substituting into the constraint $a^2 + b^2 = 3$, we get $2\cos(\alpha - \beta) = a^2 + b^2 - 2 = 1$, giving $\frac{\sin(2\theta)}{\cos\theta} = a^2 + b^2 - 3 = 0$, so the answer is $3$.
Correct Answer: 3