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Question

If tanθ=a14a, then secθtanθ equals

A
2a
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B
12a,2a
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C
12a,2a
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D
2a,12a
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Solution

The correct option is C 2a,12a
tanθ=a14a
Letk=secθtanθ

k=1+tan2θtanθ {sec2θ=1+tan2θ}

k=1+(a14a)2a+14a

k=(a+14a)2a+14a=±(a+14a)a+14a

k=2a,12a
Ans: D

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