Proving trig identities help???

Question

Prove that cotA+tan2A= cotAsec2 (using LHS=RHS)

Answer 1

cotA + tan 2A

= 1/tanA + (2tanA)/(1-tan²A)
= [ {1-tan²A}+2tan²A ] / (1-tan²A)(tanA)
= [ (1+tan²A)/(1-tan²A) ]*[ 1/tanA ] .. multiply by cos²A both N & D
= [(cos²A + sin²A)/(cos²A - sin²A)] * [1/tanA]
= 1/cos(2A) * 1/tanA
= sec(2A) cot A

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Answer 2

in this subject, i believe you're able to desire to apply the Pythagorean, Reciprocal, and Quotient Identities...so a techniques as a textbook is worried. just about, right here - take the two factors of the id at a time: Sin^2(x) Sec^2(x) could nicely be simplified to Tan^2 (x) with the aid of fact secant is a million/cosine, consequently you get Sin^2(x) / Cos^2(x). so it fairly is now Tan^2(x) = sec^2(x) -a million. in accordance to the Pythag identity, a million+tan^2(x) = Sec^2(x). that's particularly changed via subtracting 'a million' from the two factors. This sources illustrates the tip of the info, ending with: tan^2(x) = tan^2(x)

Answer 3

Hi,

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)

cos A/sin A can be replaced by cot A.

cot A(.......1......)
--------------------------- =
.....(cos² A.-..sin² A)

cos² A - sin² A can be replaced by cos 2A

cot A
---------- =
cos 2A

Since sec A = 1/cos A, replace 1/cos² A with sec² A.

cot A sec 2A

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

LHS=cosA/sinA + sin2A/cos2 A = ( cos2A cos A+sin2AsinA)/sinAcos2A
= cos(2A-A)/sinAcos2A=cosA/sinAcos2A= cotAsec2A=RHS

Answer 5

cotA + tan 2A= cotA + (sin2A/cos2A);
= cotA + (2sinAcosA/ cot2A);
=(cosA/sinA) + (2sinAcosA/2cos^2A-1);

=(2cos^3A-cosA+2sin^2AcosA)/sinA*2cos^2A-1
=cosA[2cos^2A-1+2sin^2A]/ sinA cos2A
=cosA[2-1]/sinAcos2A
=cosA*sec2A/sinA
=cotA*sec2A

hence proved

Answer 6

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

Answer 7

search "sin rule" thats the answer