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QuestionMathsClass 10

If $$cos\theta =\frac{1}{2}$$ , the value of $$\frac{2sec\theta }{1+ta{n}^{2}\theta }$$ is
(A) 2
(B) 0
(C) 1
(D) 3

(C)
4.6
4.6

Solution

Here, $$cos\theta =\frac{1}{2}$$
$$B{C}^{2}=4-1$$
$$BC=\sqrt{3}$$
$$\therefore \frac{2sec\theta }{1+ta{n}^{2}\theta }=\frac{2\times\;2}{1+(\sqrt{3}{)}^{2}}=\frac{4}{1+3}=\frac{4}{4}=1$$
$$\therefore \left(sec\theta =\frac{2}{1},tan\theta =\frac{\sqrt{3}}{1}\right)$$