Answers

  • qais
  • Content Quality
2015-12-21T22:39:21+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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)
0
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
0