ABC is a triangle such that angle C = 60°, then what is (cosA + cosB)/cos((A–B)/2) equal to?
A.2
B.√2
C.1✓ Correct
D.1/√2
Explanation
cosA + cosB = 2cos((A+B)/2)cos((A–B)/2). Since A + B + C = π and C = 60°, A + B = 120°, so (A+B)/2 = 60°. Thus cosA + cosB = 2cos60°·cos((A–B)/2) = 2·(1/2)·cos((A–B)/2) = cos((A–B)/2). Dividing by cos((A–B)/2) gives 1.
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