Math, asked by huzefa24, 1 year ago

prove that cos(90°-A).sin(90°-A)/tan(90°-A) =
 {sin}^{2} a

Answers

Answered by Anonymous
2
Here is your solution....

cos(90°-A).sin(90°-A)/tan(90°-A) = sin²A

Lhs...

Cos(90°-A).sin(90°-A)/tan(90°-A)
SinA . CosA /CotA....

SinA . (CosA.SinA)/CosA......coz CotA= CosA/SinA....
Sin²A......because CosA /CosA....cancel out....

Hence lhs = rhs.....


Hope u like it....❤
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