Question

If tan θ₁ tan θ₂ = k, then $\frac{\cos \left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)}{\cos \left({\mathrm{\theta }}_1+{\mathrm{\theta }}_2\right)}=$

(a) $\frac{1+k}{1-k}$
(b) $\frac{1-k}{1+k}$
(c) $\frac{k+1}{k-1}$
(d) $\frac{k-1}{k+1}$

Answer 1

(a) $\frac{1+k}{1-k}$


$$\begin{gathered} \frac{\cos ({\theta }_1-{\theta }_2)}{\cos ({\theta }_1+{\theta }_2)} \\ =\frac{\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2}{\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2} \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\cos {\theta }_1\cos {\theta }_2,\text{we get}: \end{gathered}$$
$$\begin{gathered} \frac{1+\tan {\theta }_1\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2} \\ =\frac{1+k}{1-k} \end{gathered}$$