# Prove that CotA-Cot2A=cosec2A

2
by yasuLivbhath

• qais
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2015-12-21T22:39:21+05:30

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LHS
cot A - cot 2A = (cosA/sinA)- (cos2A/sin2A)
=(cosA sin2A -sinA cos2A)/(sinA sin2A)
∵ cos2A =cos²A - sin²A
and, sin2A= 2sinA cosA
putting the values,
(2sinA cos²A -sinA (cos²A -sin²A))/(2sin²A cosA)
taking common sinA and cancelling sinA ,
(cos²A + sin²A)/(2sinA cosA)
=1/sin2A
= cosec2A = RHS   (proved)
2015-12-22T12:45:08+05:30
The question you have given is wrong . Its cot A - cot 2A = 2cosec 2A .

There is also another fourmula

cot A + cot 2A = 2 tan 2A

Proof :

cot A - cot 2A = 1/tan A - 1/tan 2A

tan 2A = 2tan A / 1 - tan^2 A

So

1/tan A - (1 - tan^2 A ) / 2tan A

= 1/tan A ( 2 - 1 + tan^2 A ) = 1 + tan^2 A / tan A

= 2 * 1/sin 2A = 2 cosec 2A

since sin 2A = 2tan A / 1 + tan^2 A