Proving trig identities help?

Question

cotA+tan2A= cotAsec2A

Answer 1

LHS
cos A / sin A + sin 2A / cos 2A
(cos A cos 2A + sin A sin 2A) / (sin A cos 2A)
cos (A - 2A) / (sin A cos 2A)
cos (- A) / (sin A cos 2A)
cos A / (sin A cos 2A)
(cos A / sin A) (1 / cos 2A)
cot A sec 2A

RHS
cot A sec 2A

LHS = RHS

Answer 2

If it is an identity then it will be tru for all values.
Let the value of A be 30 degrees
Therefore,
=>CotA + Tan2A = CotASec2A
=>Cot30 + Tan (2*30) = Cot30*Sec(2*30)
=>Cot30 + Tan60 =Cot30.Sec60
=>sqrt. 3 + sqrt. 3 = sqrt. 3 * 2
=>2sqrt.3 = 2sqrt. 3
=>Hence LHS =RHS
Hence Proved.

Answer 3

Hi,

The right side can be re-written as:

cot A sec 2A =

cos A............1
---------.x.--------------------
sin A.......cos² A - sin² A

cos A
-----------------------------
sin A(cos² A - sin² A)


The left side can be changed into:

cot A + tan 2A =

cos A /sin A + (2 tan A)/(1 - tan² A) =

cos A /sin A + (2 sin A/cos A)/(1 - sin² A/cos² A)

Multiply this last fraction on its top and bottom by cos² A to eliminate smaller fractions.

2sin A * cos² A
---------
cos A
========
(........sin² A.)cos² A
(1.-...---------)
(........cos² A)

2sin A cos A
--------------------
cos² A.-..sin² A

So now the left side is:

cos A.......2sin A cos A
---------.+.---------------------
sin A.......cos² A.-..sin² A

Getting a common denominator gives:

cos A(cos² A.-..sin² A) + 2sin A cos A*sin A
------------------------------------------------------------
sin A(cos² A.-..sin² A)


cos³ A.-..cos A sin² A + 2cos A sin² A
----------------------------------------------------
sin A(cos² A.-..sin² A)

cos³ A.+..cos A sin² A
-------------------------------
sin A(cos² A.-..sin² A)

Factor out cos A in the numerator. Remember cos² + sin² = 1.

cos A(cos² A.+.sin² A)
-------------------------------
sin A(cos² A.-..sin² A)

cos A(.......1......)
-------------------------------
sin A(cos² A.-..sin² A)

This now matches the right side because cos A/sin A = cot A.
cos² A.-..sin² A is the same as cos 2A, but since it's really 1/cos 2A, that equals sec 2A by the reciprocal identities.

I hope that helps!! :-)

Answer 4

first of all sec2A = sec^2A/(1-tan^2A) (a)

cotA+tan2A=cotA+(2tanA/1-tan^2a)
=1/tanA(1+<2 tan^2 A/1-tan^2A>)
=cotA(1+tan^2A/1-tan^2a)
=sec^2A/) (b)

using (a) and(b)
lhs=sec2Acota

Answer 5

cotA+tan2A= cotA*sec2A

cosA/sinA + sin2A/cos2A = (cosA/ sinA ) * 1 / cos2A
(cosA*cos2A + sin2A*sinA) / (sinA * cos2A) =
cosA / (sinA / cos2A)

Remove sinA * cos2A in both denominators:
cosA*cos2A + sin2A*sinA = cosA
cosA*cos2A + 2sinA*cosA*sinA = cosA

Divide per cosA:
cos2A + 2sinA*sinA = 1
cos2A + 2sin²A= 1

cos²A - sin²A + 2sin²A = 1
cos²A + sin²A = 1
1 = 1

Answer 6

(cosA/sinA)+(sin(2A)/cos(2A))
taking LCM
( cosAcos(2A)+sinAsin(2A) )/( sinAcos(2A) )
cos(a-b)=cosa cosb +sina sinb
a=A b=2A
cos(A-2A)/( sinAcos(2A) )
cos(-theta)=cos(theta)
cos(A)/( sinAcos(2A) )
=(cosA/sinA)*(1/cos(2A) )
=cotA*sec(2A)