Prove that cos 105° + cos 15° = sin 75° − sin 15°.

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Solution

LHS = cos105^o + cos15^o
= cos(90^o + 15^o) + cos(90^o $-$ 75^o)
= - sin 15^o + sin 75^o [As cos(90^o+A) = $-$ sin A and cos(90^o $-$ B) = sin B]
= sin 75^o $-$ sin 15^o
= RHS
Hence proved.

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