Question

tanO+sinO=m and tanO-sinO=n.Show that m^2-n^2=4rootmn​

Answer 1

Answer:

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

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Answer 2

Answer:

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