Question

prove that cos square A minus 1 cos square A + 1 equal to -1​

Answer 1

Answer:

${ \cos}^{2} a - 1 { \cos }^{2} a + 1 = - 1 \\ \frac{1}{ { \sin}^{2}a } - 1 \times \frac{1}{? \ { \cos}^{2}a } + \frac{1}{1} \\ \frac{1 - \ \ { \sin}^{2}a }{\ { \sin }^{2}a } \times \frac{1 + { \cos }^{2}a }{? { \cos }^{2} a} = - 1 \\ \frac{ { \ - cos}^{2} a }{ { \sin(?) }^{2} a} \times \frac{ { \ \sin }^{2} a}{ { \cos }^{2} a} = - 1 \\ = - 1 = - 1$

hence proved

Answer 2

As we know tan A=Sin A/Cos A and tan B=Sin B/Cos B  So

tan squareA - tan squareB= (Sin A/Cos A)^2 - (Sin B/Cos B)^2 =

{(Sin^2 A Cos^2 B)-(Cos^2 A Sin^2 B)}/(Cos^2 A .Cos^2 B)

As we know that Sin^2 A=1-Cos^2 A and Sin^2 B=1-Cos^2 B ,So

tan squareA - tan squareB={( Cos^2 B(1-cos^2 A))-(Cos^2 A (1-Cos^2 B))}/(Cos^2 A .Cos^2 B)= (Cos^2 B- Cos^2 A)/(Cos^2 A .Cos^2 B)

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