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If m=tanA+sinA and m 2 -n 2 =4√mn Then show that, n=tanA-sinA

If m=tanA+sinA and m2-n2=4√mn
Then show that,
n=tanA-sinA

Grade:10

2 Answers

Arun
25750 Points
4 years ago
let's consider the expression:  
(tanA + sinA)² - (tanA - sinA)² =  
let's factor it as a difference of squares:  
[(tanA + sinA) + (tanA - sinA)][(tanA + sinA) - (tanA - sinA)] =  
(tanA + sinA + tanA - sinA)(tanA + sinA - tanA - sinA) =  
(2tanA)(2sinA) =  
4tanA sinA =  
(writing tanA as sinA /cosA)  
4(sinA/cosA) sinA =  
4(sin²A /cosA)  
we can rewrite this (on condition that cosA is positive) as:  
4√(sin²A /cosA)² =  
4√[(sin²A)² /cos²A] =  
let's rewrite the numerator of the radicand as:  
4√{[(sin²A)(sin²A)] /cos²A} =  
let's replace the second factor in the numerator with 1 - cos²A:  
4√{[sin²A (1 - cos²A)] /cos²A} =  
4√[(sin²A /cos²A)(1 - cos²A)]} =  
(expanding and simplifying)  
4√[(sin²A /cos²A) - (sin²A /cos²A) cos²A] =  
4√[(sinA /cosA)² - sin²A] =  
4√(tan²A - sin²A) =  
(factoring the radicand as a difference of squares)  
4√[(tanA + sinA)(tanA - sinA)]  
summing up, we have:  
(tanA + sinA)² - (tanA - sinA)² = 4√[(tanA + sinA)(tanA - sinA)]  
where tanA + sinA = m:  
m² - (tanA - sinA)² = 4√[m (tanA - sinA)]  
on the other hand, we know that:  
m² - n² = 4√(m n)  
then, comparing the two expressions, we conclude that:  
tanA - sinA = n  
 
Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
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