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if tanA+sinA=m, and tanA-sinA=n, show that m2-n 2 = 4 under root 'mn'

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4 under root mn

4 under root (tanA + sinA ) (tan A - sinA )

4 " " tan2A - sin2A

4 " " sin2A/ cos2A - sin2A

4 " "  sin2A - sin2A . cos2A / (whole by) cos2A

4 " " (taking sin2A out ) sin2A ( 1-cos2A) / cos2A

4 " "  sin4A/ cos2A

4 x sin2A /cosA             (taking square out to remove "root")

4 x sinA x sinA/cosA

4xsinAxtanA

Now taking LHS = m2 - n2

(tanA + sinA)2 - (tanA -sinA )2

Since   (a + b )2 - (a -b)2 = 4ab

4xsinAxtanA  so LHS = RHS

  • 42

m=tanA+sinA

n=tanA-sinA

L.H.S= ((tanA+sinA)2 - (tanA-sinA)2)2

(tan2A+sin2A+2tanAsinA -(tan2A+sin2A-2tanAsinA))2

(tan2A + sin2A + 2tanAsinA - tan2A - sin2A + 2tanAsinA)2

(4tanAsinA)

16 tan2Asin2A

16 tan2A(1-cos2A)

16 tan2A - tan2Acos2A

16 tan2A - sin2A/cos2A x cos2A

16 tan2A - sin2A

16 (tan2A +sin2A) (tan2A - sin2A)

16mn

   OR

4 root mn

  • 16
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