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

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Grade 10Trigonometry

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

Tawsif Ahmed

6 Years agoGrade 10

2 Answers

Arun

6 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