If cos$$\theta$$ =$$\frac{P}{\sqrt{p^{2}+q^{2}}}\ $$, then the value of tan$$\theta\ $$ is:

Expression : cos$$\theta$$ =$$\frac{P}{\sqrt{p^{2}+q^{2}}}\ $$ -------------(i)

=> $$\frac{1}{cos\theta}=\frac{\sqrt{p^2+q^2}}{p}$$

=> $$sec\theta=\frac{\sqrt{p^2+q^2}}{p}$$

Squaring both sides, we get :

=> $$sec^2\theta=(\frac{\sqrt{p^2+q^2}}{p})^2$$

=> $$sec^2\theta=\frac{p^2+q^2}{p^2}$$

Subtracting '1' from both sides,

=> $$sec^2\theta-1=\frac{p^2+q^2}{p^2}-1$$

=> $$tan^2\theta=\frac{(p^2+q^2)-p^2}{p^2}$$

=> $$tan^2\theta=\frac{q^2}{p^2}$$

Taking square root on both sides, we get :

=> $$\sqrt{tan^2\theta}=\sqrt{\frac{q^2}{p^2}}$$

=> $$tan\theta=\frac{q}{p}$$

=> Ans - (A)