Question #ca9fe

2 Answers
P dilip_k
Apr 4, 2017

#LHS=cscA*sin2A-secA#

#=cscA*(2sinA*cosA)-secA#

#=1/sinA*(2sinAcosA)-1/cosA#

#=2cosA-1/cosA#

#=(2cos^2A-1)/cosA#

#=(cos2A)/cosA#

#=cos2A*secA=RHS#

Hugge
Apr 4, 2017

Use trigonometric identities as done below

Explanation:

We will make use of the identities below, found here and here . As for why we choose these identities, I can only refer to practice and intuition.
1. #csc(x) = 1/sin(x)#
2. #sec(x) = 1/cos(x)#
3. #sin(2x) = 2sin(x)cos(x)#
4. #cos(2x) = cos^2(x) - sin^2(x)#
5. #sin^2(x) + cos^2(x) = 1#

Rewrite #csc(A)sin(2A)-sec(A)# using identies 1. and 2.
#csc(A)sin(2A)-sec(A) = #
#sin(2A)/sin(A)-1/cos(A)#
use identity 3.
#(2sin(A)cos(A))/sin(A)-1/cos(A)#
#2cos(A) - 1/cos(A)#
Put on common denominator by multiplying the first term by #cos(A)#
#(2cos^2(A) - 1)/cos(A)#
Substitute the #1#, using identity 5.
#(2cos^2(A) - (cos^2(A) + sin^2(A)))/cos(A)#
Simplify (and mind the sign when removing the parenthesis)
#(cos^2(A) - sin^2(A))/cos(A)#
Use identity 4 and write the denominator as a factor, for clarity
#cos(2A)1/cos(A)#
Use identity 2.

#cos(2A)sec(A)#

and we have arrived at what we wanted to show.

Related questions
See all questions in Proving Identities
Impact of this question
1318 views around the world
You can reuse this answer
Creative Commons License