If tan A = n tan B and sin A = m sin B , then prove that cos ² A = m²-1 / n²-1.
(Class 10 Maths Sample Question Paper)
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Answered by
576
Solution:
In this question we have to find cos²A in terms of m and n , so we have to eliminate ∠B from the given relations.
tan A = n tan B
tan B = 1/n tan A
Cot B = n /tan A [ cot B = 1/tan B]
sin A = m sinB
sin B = 1/m sinA
cosec B = m / sinA [sinB = 1/cosecB]
cosec²A - cot²B =1
Substitute the value of cot B and cosec B in the above relation.
(m / sinA)² - (n /tan A)²
(m² / sin²A) - (n² /tan² A)
(m² / sin²A) - (n² /(sin²A / cos²A))
[ tan A = sinA / cosA]
(m² / sin²A) - n²cos²A / sin²A = 1
m² - n²cos²A = sin²A
m² - n²cos²A = 1- cos²A
[sin²A = 1- cos²A]
m² -1 = n²cos²A - cos²A
m² - 1 = cos ²A(n² -1)
cos²A = m² -1/ n²-1
HOPE THIS WILL HELP YOU...
In this question we have to find cos²A in terms of m and n , so we have to eliminate ∠B from the given relations.
tan A = n tan B
tan B = 1/n tan A
Cot B = n /tan A [ cot B = 1/tan B]
sin A = m sinB
sin B = 1/m sinA
cosec B = m / sinA [sinB = 1/cosecB]
cosec²A - cot²B =1
Substitute the value of cot B and cosec B in the above relation.
(m / sinA)² - (n /tan A)²
(m² / sin²A) - (n² /tan² A)
(m² / sin²A) - (n² /(sin²A / cos²A))
[ tan A = sinA / cosA]
(m² / sin²A) - n²cos²A / sin²A = 1
m² - n²cos²A = sin²A
m² - n²cos²A = 1- cos²A
[sin²A = 1- cos²A]
m² -1 = n²cos²A - cos²A
m² - 1 = cos ²A(n² -1)
cos²A = m² -1/ n²-1
HOPE THIS WILL HELP YOU...
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417
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