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Prove that sec2θ+cosec2θ can never be less than 2.
Ans: S.T Sec2θ + Cosec2θ can never be less than 2.
If possible let it be less than 2.
1 + Tan2θ + 1 + Cot2θ < 2.
⇒ 2 + Tan2θ + Cot2θ
⇒ (Tanθ + Cotθ)2 < 2.
Which is not possible.
In figure, the incircle of triangle ABC touches the sides BC, CA, and AB at D, E, and F respectively. Show that AF+BD+CE=AE+BF+CD= 1/2 (perimeter of triangle ABC), Ans:
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give me the anwsers ..
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