Question

Prove the following:
cos (90° - A) sin (90° - A)/
tan (90° - A)
= 1-COS^2 A​

Answer 1

L.H.S.

$\frac{cos (90° - A) sin (90° - A)}{tan (90° - A)} \\ = > \frac{sinAcosA}{cot A } \\ = > \frac{sinAcosA}{ \frac{cosA}{sinA} } \\ = > sinAcosA \times \frac{sinA}{cosA} \\ = > sin ^{2} A \\ = > 1 - cos^{2} A$

= RHS

Hence, proved.

Remember:

cos(90°-A)=SinA

Sin(90°-A)=cosA

tan(90°-A)=cotA

cotA= cosA/sinA

sin²A+cos²A=1 => sin²A=1-cos²A