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.

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