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Question
lf A + B + C = π, then `cosA/(sinBsinC) + cosB/(sinCsinA) + cosC/(sinAsinB)` is equal to ______.
Options
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MCQ
Fill in the Blanks
Solution
lf A + B + C = π, then `cosA/(sinBsinC) + cosB/(sinCsinA) + cosC/(sinAsinB)` is equal to 2.
Explanation:
LHS = `1/2 (sin 2A + sin 2B + sin 2C)/(sin A sin B sin C)` = 2
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Factorization Formulae - Trigonometric Functions of Angles of a Triangle
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