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Question

If tan2x+secxa=0 has alteast one solution, then a..........


A

(,1]

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B

[-1,1]

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C

[1,)

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D

(,1]

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Solution

The correct option is B

[-1,1]


tan2x+secxa=0(sec2x1)+secxa=0=sec2x+secx+141a=0=(secx+12)=54+aNowsecx1or1secx+1232orsecx+1212(Secx+12)294or(Secx+12)214i.e.54+a14=a1


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