Question

Hello dears

© Show that :

1 - cos2∅ + sin2Ø / 1 + cos2∅ + sin2∅ = tan∅

© show that :

sinAcosA + sinBcosB/ sin²A - sin²B = cos ( A - B )

© show that :

1 + sin2A / cosA = cosA + sinA/ cosA + sinA = tanA ( 45° - A )

points : 50

thanks

expected answers from special users..

Answer 1

L.H.S

1- (1-2sin²∅)+2sin∅cos∅/ 1 + (2cos²∅-1) + 2sin∅cos∅

=2sin²∅+2sin∅cos∅/2cos²∅+ 2sin∅cos∅

=2sin∅(sin∅+cos∅)/2cos∅(cos∅+sin∅)

=sin∅/cos∅

=tan∅

=RHS

2) sinAcosA + sinBcosB/ sin²A - sin²B = Cot( A - B )

sinAcosA+sinBcosB/ sin²A-sin²B *2/2

= sin2A+sin2B/2 sin²A-sin²B

=2sin(2A+2B/2)cos(2A-2B/2) / 2sin(A+B)sin(A-B)

=2sin(A+B)cos(A-B)/ 2sin(A+B)sin(A-B)
=cos(A-B)/sin(A-B)
=cot (A-B)

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer ;}}}$

• See the attached file ⤴️