Question

cos2A+cos2B-cos2c=1-4sinA sinB cosc​

Answer 1

Answer:

L.H.S.

=2.cos(2A+2B)/2.cos (2A-2C)/2-(2cos^2C-1).

= 2.cos(180°-C).cos(A-B)-2cos^2 C. +1.

= - 2.cosC.cos(A-B)-2.cos^2 C. +1

= 1 -2cosC[cos(A-B) +cosC].

= 1 -2.cosC[ cos(A-B)+cos{180°-(A+B)}]

= 1–2cosC.[cos(A-B)-cos(A+B)]

= 1–2.cosC.[2sinA.sinB ]

= 1. - 4.sinA.sinB.cosC. Proved.

Step-by-step explanation:

hope it helps:-)

Answer 2

Step-by-step explanation:

L.H.S.

=2.cos(2A+2B)/2.cos (2A-2C)/2-(2cos^2C-1).

= 2.cos(180°-C).cos(A-B)-2cos^2 C. +1.

= - 2.cosC.cos(A-B)-2.cos^2 C. +1

= 1 -2cosC[cos(A-B) +cosC].

= 1 -2.cosC[ cos(A-B)+cos{180°-(A+B)}]

= 1–2cosC.[cos(A-B)-cos(A+B)]

= 1–2.cosC.[2sinA.sinB ]

= 1. - 4.sinA.sinB.cosC.

R.H.S

=1. - 4.sinA.sinB.cosC

HENCE PROOVED L.H.S=R.H.S.