Question

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Answer 1

Answer:

Given as cos 55° + cos 65° + cos 175° = 0 Let us consider the LHS cos 55° + cos 65° + cos 175° On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 cos 55° + cos 65° + cos 175° = 2 cos (55° + 65°)/2 cos (55° – 65°) + cos (180° – 5°) = 2 cos 120°/2 cos (-10°)/2 – cos 5° (since, {cos (180° – A) = – cos A}) = 2 cos 60° cos (-5°) – cos 5° (since, {cos (-A) = cos A}) = 2 × 1/2 × cos 5° – cos 5° = 0 = RHS Thus proved.