question_answer

If \[\cos ec\,\theta =\sqrt{10},\]then \[\sec \,\theta =\]

A) \[\frac{3}{\sqrt{10}}\]

B) \[\frac{\sqrt{10}}{3}\]

C) \[\frac{1}{\sqrt{10}}\]

D) \[\frac{2}{\sqrt{10}}\]

Correct Answer: B

Solution :

[b] Let \[\Delta ABC\] is right triangle such that \[\angle B=90{}^\circ \] and \[\angle A=\theta \].

\[\therefore \,\,\,\,\,\,\cos ec\,\theta =\frac{AC}{BC}=\frac{\sqrt{10}}{1}\]

Let \[AC=\sqrt{10}k\] and \[BC=k,\]

Where, k is positive constant.

\[\therefore \,\,\,\,A{{B}^{2}}=(A{{C}^{2}}-B{{C}^{2}})=(10{{k}^{2}}-k\,)=9{{k}^{2}}\]

\[\Rightarrow \,\,\,\,\,\,\,AB=\sqrt{9{{k}^{2}}}=3k\]

\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \,\theta =\frac{AC}{AB}=\frac{\sqrt{10}k}{3k}=\frac{\sqrt{10}}{3}\]