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Find the value of cot2θ−cos2θ\sqrt{cot^2\theta-cos^2\theta}.

Acot θ cosec θ

B1

Ccos θ cosec θ

Dcot θ cos θ

Answer:

D. cot θ cos θ

Read Explanation:

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In the given figure ∠ABC=∠ABD,BC=BDthen△CAB=△\angle{ABC} = \angle{ABD}, BC = BD then \triangle{CAB} =\triangle___________

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