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If A + B + C = 180°, prove that sin2A + sin2B − sin2C = 2 sin A sin B cos C

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प्रश्न

If A + B + C = 180°, prove that sin2A + sin2B − sin2C = 2 sin A sin B cos C

आलेख
बेरीज
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उत्तर

LH.S = `(1 - cos2"A")/2 + (1 - cos2"B")/2 - (1 - cos2"C")/2`

Hint: `[sin^2"A" = (1 - cos2"A")/2]`

= `[1/2 + 1/2 -  1/2] - 1/2 [cos2"A" + cos2"B" - cos 2"C"]`

= `1/2 - 1/2 [2cos("A" + "B") cos("A" - "B") - (2cos^2"C" - 1)]`

= `1/2 - cos("A" + "B") cos("A" - "B") + cos^2"C" - 1/2`

= cos C cos(A – B) + cos2C

= cos C [cos(A – B) – cos(A + B)]

= cos C [2 sin A sin B]

= 2 sin A sin B cos C

= R.H.S

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Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.7 [पृष्ठ १२४]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.7 | Q 1. (iv) | पृष्ठ १२४

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