can you solve that cot2A+TanA=Cosec2A?

Answer 1

cot 2A + tan A

= cos 2A/sin 2A + sin A/cos A

= (cos 2A cos A + sin 2A sin A) / sin 2A cos A

Using the compound angle formulae,

we know that cos (2A - A ) = cos 2A cos A + sin 2A sin A...

therefore,

= cos (2A - A) / sin 2A cos A
= cos A / sin 2A cos A
= 1/ sin 2A
= cosec 2A ( proven)

Good luck !!!

Answer 2

If I remember my trig equations correctly, this equation is true for all real numbers A.

cot 2A = cos 2A / sin 2A = (cos^2 A - sin^2 A)/(2 sin A cos A)
csc 2A = 1/sin 2A = 1/(2 sin A cos A)

So, rewritten in terms of sin A and cos A, the equation becomes

(cos^2 A - sin^2 A)/(2 sin A cos A) + (sin A / cos A) = 1/(2 sin A cos A)

Multiplying the entire equation by 2 sin A cos A:

(cos^2 A - sin^2 A) + 2 sin^2 A = 1
cos^2 A + sin^2 A = 1

Which is true for all real A.

Answer 3

tan A= cosec2A-cot2A
= (1-cos2A)/sin2A = 2sin^2 A/(2sinAcosA)
= sinA/cosA = tanA
hence, solns r infinite, A {- R ( A belongs to set f real nos)
A cn b nething

Answer 4

giben cos 2a/sin 2a + sin a/cos a = 1/ sin 2a
multiply by sin 2a

cos 2a + sin a sin 2a/cos a = 1

or cos 2a + 2 sin^ a = 1 as sin 2a/cos a = 2 sin a

or cos 2a = 1- 2 sin ^a = 1- sin^a - sin^a = cos^a - sin^a

which is true
because cos(a+b) = cos a cos b - sina sin b
put b = a
cos 2a = cos^a - sin ^a

this is true for all a so proved

Answer 5

Solved using numerical technique:
newton ralphson method to get
A=1.03064516 radians

Answer 6

Your question must begin with prove that...
or show that not " can you solve"
instead of the word "solve",
you should use "show"
Because this an identy not equation to solve.

LHS=cos 2A/sin2A +sinA/cosA
=(cos2AcosA+sin2A sinA)/(cosA . sin2A)
=cos(2A-A)/(cosAsin2A)
=cosA/(cosA sin2A)
=1/sin2A

RHS=1/sin2A

LHS=RHS
So, the given identity is true.

Answer 7

If your Question is cot2A+tan2A=Cosec2A then i can answer.

Answer 8

I think the question is:
Cot2A + Tan 2A = Cosec 2A.

Answer 9

cot(2A) + tanA = csc(2A)

Assuming you don't mean cot(A)^2 + tanA = csc(A)^2

this is the same as saying

(1/(tan(2A))) + tanA = 1/(sin(2A))

becomes

(1 + (tan(A) * tan(2A))/(tan(2A))) = 1/(sin(2A))

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

(1 + (tan(A) * ((2tanA)/(1 - tan(A)^2)))/((2tan(A))/(1 - tan(A)^2))

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

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

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

(1 + tan(A)^2)/(1 - tan(A)^2) * ((1 - tan(A)^2)/(2tanA))

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

(1 + tan(A)^2)/(2tanA)

tan(A)^2 + 1 = sec(A)^2

sec(A)^2 = 1/(cos(A)^2)

(1/(cos(A)^2)) / (2(sinA/cosA))

(1/(cos(A)^2)) / ((2sinA)/(cosA))

(1/(cos(A)^2)) * ((cosA)/(2sinA))

(cosA)/(2sinAcosA^2)

1/(2sinAcosA)

1/(2sinAcosA) = 1/(sin(2A))

1/(sin(2A)) = csc(2A)

so

cot(2A) + tanA = csc(2A)

Answer 10

Yes, now the real question is can you?