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If A + B + C = 180°, then sin2A+sin2B+sin2CcosA+cosB+cosC-1 is equal to ______. -

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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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