if tanA + sinA = m and tanA - sinA= n prove that (m²-n²)=16
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anA +sinA = m,
TanA- sinA = n,
Now LHS,
m² - n²
(tanA + sinA)² - (tanA - sinA)²
(tan²A + sin² A + 2tanAsinA ) - ( tan²A + sin²A - 2tanAsinA)
tan²A + sin²A +2 tanA sinA -tan²A -sin²A +2tanA sinA
4tanA sinA ______ (1)
Now RHS,
16mn
16(tanA + sinA) ( tanA - sinA)
16( tan²A - sin² A) _______[a² - b² = ( a+ b ) (a- b)]
16 tan²A ( 1-cos²A)
16 tan²A sin²A
(4 tanA sinA)²
Here we can see LHS is not equal to RHS.
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