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If A + B + C = 180°, prove that sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4 sin A sin B sin C

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प्रश्न

If A + B + C = 180°, prove that sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4 sin A sin B sin C

योग
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उत्तर

Now A + B + C = 180°

So B + C = 180° – A

sin(B + C – A) = sin(180° – A – A)

= sin(180° – 2A) = sin 2A

Now L.H.S = sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C  ......[From (i)]

= R.H.S

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.7 [पृष्ठ १२४]

APPEARS IN

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

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