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Question

If tan θ=3, thensec2θ-cosec2θsec2θ+cosec2θ=?
(a) −1
(b) 1
(c) -12
(d) 12

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Solution

(d) 12
Here,
tanθ=3
ACBC=31
By Pythagoras' theorem, we get AB = 1.

Now, secθ=ABBC=21 and cosecθ=ABAC=23
sec2θ-cosec2θsec2θ+cosec2θ=22-23222+232=4-434+43=12-412+4=816=12

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