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Q.

If tan2θ=1−k2 then secθ+tan3θcosecθ=

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a

2−k2

b

2+k2

c

(2−k2)32

d

(2+k2)32

answer is C.

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Detailed Solution

1cosθ+sin3θcos3θ×cosecθ=cos2θ+sin2θcos3θ=sec3θ→1 Given tan2θ=1-k2, ⇒sec2θ=2-k2, ⇒secθ =2-k2, ⇒sec3θ=2-k232→2 By 1 and 2, we have 1cosθ+sin3θcos3θ×cosecθ=2-k232

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