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Grade:

                        

Solve the following question Prove that- b² cos2A - a² cos2B = b²-a²

2 years ago

Answers : (1)

Arun
24741 Points
							
Dear Sanjay
 
First use the formula cos 2A = 1 - 2sin²A 

(1 - 2sin²A)/a² - (1 - 2sin²B)/b² = 1/a² - 2sin²A/a² - 1/b² + 2sin²B/b² 

= 1/a² - 1/b² + 2sin²B/b² - 2sin²A/a² 

Now think of the law of sines in a triangle 

sinA/a = sinB/b 
sin²A/a² = sin²B/b² 
Hence 
2sin²B/b² - 2sin²A/a² = 0 

So 

Cos2A/a² - cos2B/b² = 1/a² - 1/b² 
2 years ago
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