question_answer

If \[\cos \theta =\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}},\]then the value of \[\cot \theta \] is equal to \[[If\,0\le \,\,\theta \le \,90{}^\circ ]\][SSC CGL Tier II, 2017]

A) \[\frac{2xy}{{{x}^{2}}-{{y}^{2}}}\]

B) \[\frac{2xy}{{{x}^{2}}+{{y}^{2}}}\]

C) \[\frac{{{x}^{2}}+{{y}^{2}}}{2xy}\]

D) \[\frac{{{x}^{2}}-{{y}^{2}}}{2xy}\]

Correct Answer: A

Solution :

[a] Given, \[\cos \theta =\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\]                         In right angled \[\Delta ABC,\]\[{{(BC)}^{2}}={{(AC)}^{2}}-{{(AB)}^{2}}\] \[{{(BC)}^{2}}=[{{({{x}^{2}}+{{y}^{2}})}^{2}}-{{({{x}^{2}}-{{y}^{2}})}^{2}}]\] \[{{(BC)}^{2}}=({{x}^{2}}+{{y}^{2}}+{{x}^{2}}-{{y}^{2}})({{x}^{2}}+{{y}^{2}}-{{x}^{2}}+{{y}^{2}})\] \[{{(BC)}^{2}}=2{{x}^{2}}\times 2{{y}^{2}}=4{{x}^{1}}{{y}^{2}},\] \[\therefore \]\[\cot \theta =\frac{AB}{BC}=\frac{{{x}^{2}}-{{y}^{2}}}{2xy}\]