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Question
If A + B + C = 180°, then `(sin 2A + sin 2B + sin 2C)/(cos A + cos B + cos C - 1)` is equal to ______.
Options
`8 sin A/2 sin B/2 sin C/2`
`8 cos A/2 cos B/2 cos C/2`
`8 sin A/2 cos B/2 cos C/2`
`8 cos A/2 sin B/2 sin C/2`
MCQ
Fill in the Blanks
Solution
If A + B + C = 180°, then `(sin 2A + sin 2B + sin 2C)/(cos A + cos B + cos C - 1)` is equal to `underlinebb(8 cos A/2 cos B/2 cos C/2)`.
Explanation:
We know that,
sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
and cos A + cos B + cos C – 1 = `4 sin A/2 sin B/2 sin C/2`
Hence, `(4 sin A sin B sin C)/(4 sin A/2 sin B/2 sin C/2)`
= `(4.2 sin A/2 cos A/2 . 2 sin B/2 cos B/2 . 2 sin C/2 cos C/2)/(4 sin A/2 sin B/2 sin C/2)`
= `8 cos A/2 cos B/2 cos C/2`
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Factorization Formulae - Trigonometric Functions of Angles of a Triangle
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