Question

Cos105°+cos15° =sin75°-sin15°

Answer 1

LHS = cos105° +cos15°

=cos(90°+15°)+cos(90°-75°)

/* We know that,

i ) cos(90°+A) = -sinA

ii) cos(90°-A) = SinA

= -sin15°+sin75°

=sin75°-sin15°

=RHS

Therefore,

cos105°+cos15° = sin75°-sin105°

hope it helps

mark the brainliest

Answer 2

Answer:

Step-by-step explanation:

Cos105° + Cos 15°

= Cos(90 + 15°) + Cos(90 - 15°)

= -Sin15° + Sin75°   (∵ Cos(90 - θ) = Sinθ; Cos(90+θ) = -Sinθ)

= R.H.S

Hence proved