MCQ
If $\tan\theta_1\tan\theta_2=\text{k},$ then $\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=$
- ✓$\frac{1+\text{k}}{1-\text{k}}$
- B$\frac{1-\text{k}}{1+\text{k}}$
- C$\frac{\text{k}+1}{\text{k}-1}$
- D$\frac{\text{k}-1}{\text{k}+1}$
✓
Answer
Correct option: A.
$\frac{1+\text{k}}{1-\text{k}}$
$\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}$
$=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-0\sin\theta_1\sin\theta_2}$
Dividing numerator and denominator by $\cos\theta_1\cos\theta_2,$ we get:
$\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}$
$=\frac{1+\text{k}}{1-\text{k}}$
$=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-0\sin\theta_1\sin\theta_2}$
Dividing numerator and denominator by $\cos\theta_1\cos\theta_2,$ we get:
$\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}$
$=\frac{1+\text{k}}{1-\text{k}}$
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