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TanA+sinA=m and tanA_sinA=n show that (m^2-n^2)=16mn

TanA+sinA=m and tanA_sinA=n show that (m^2-n^2)=16mn

Grade:10

2 Answers

Arun
25750 Points
5 years ago
m = tanA + sinA
n = tanA - sinA
L.H.S. = {(tanA + sinA)² - (tanA-sinA)²}
{tan²A + sin²A + 2tanAsinA - (tan²A + sin²A - 2tanAsinA)}²
⇒ = (4tanAsinA)²
16tan²Asin²A
16tan²A(1 - cos²A)
16tan²A - tan²Acos²A
16tan²A - sin²A/cos²A × cos²A
16tan²A - sin²A
16(tan²A + sin²A) (tan²A - sin²A)
= 16 mn   
Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
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