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Q.

Let S=θ0,π2 : m=19secθ+(m1)π6secθ+6=83. Then

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a

θSθ=π2

b

S=π12

c

S=2π3

d

θSθ=3π4

answer is C.

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Detailed Solution

Let α=θ+(m1)π6

&  β=θ+mπ6

So, β-α=π6

Here, m=19secαsecβ=m=191cosαcosβ

              =2m=19sin(βα)cosαcosβ=2m=19(tanβtanα)=2m=19tanθ+mπ6tanθ+(m1)π6=2tanθ+9π6tanθ=2(cotθtanθ)=83

(Given)

 tanθ+cotθ=43

   tanθ=13

So,  S=π6,π3

θSθ=π6+π3=π2

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