MCQ
If $\tan\theta_1\tan\theta_2=\text{k},$ then $\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=$
  • A
    $\frac{1+\text{k}}{1-\text{k}}$
  • 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

  1. $\frac{1+\text{k}}{1-\text{k}}$

Solution:

$\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}$

$=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-0\sin\theta_1\sin\theta_2}$

Dividing numerator and denominator by $\cos\theta_1\cos\theta_2,$ we get:

$\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}$

$=\frac{1+\text{k}}{1-\text{k}}$

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