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Q.

tan2α=1−p2, then secα+tan3αcosecα=

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a

(2+p2)32

b

(1+p2)32

c

(2−p2)32

d

(1−p2)32

answer is C.

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Detailed Solution

secα1+sin3αcos3αcosαsinα=secα[1+tan2α]=sec3α=(sec2α)3/2=1+tan2α 32=(2−P2)3/2
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tan2α=1−p2, then secα+tan3αcosecα=