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Prove that cos(A + B) cos(A – B) = cos2A – sin2B = cos2B – sin2A

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प्रश्न

Prove that cos(A + B) cos(A – B) = cos2A – sin2B = cos2B – sin2A

योग
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उत्तर

L.H.S = cos(A + B) cos(A – B)

= (cos A cos B – sin A sin B)(cos A cos B + sin (A sin B)

= cos2A cos2B – sin2A sin2B

= cos2A (1 – sin2B) – (1 – cos2A) sin2B

= cos2A – cos2A sin2B – sin2B + cos2A sin2B

= cos2A – sin2

= R.H.S

Now cos2A – sin2B = (1 – sin2A) – (1 – cos2B)

= 1 – sin2A – 1 + cos2B

= cos2B – sin2A

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.4 [पृष्ठ १०९]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.4 | Q 17. (ii) | पृष्ठ १०९

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