α is the fourth quadrant angle.

= √{ 1/2+ 1/2 √[ 1/2 *( 1+cos 2α)]}

= √{ 1/2+ 1/2 √[ 1/2 * 2(cosα)^2]}

= √[ 1/2+ 1/2√(cosα)^2]

=√[ 1/2+ 1/2 (cosα )]

=√[ 1/2*( 1+ cosα )]

=√[ 1/2*2( cosα/2 )^2]

=√( cosα/2 )^2

=l cosα/2l

=-cosα/2

Six: left = [2sin2x * cos2x/2 (cos2x) 2] * [cos2x/2 (cosx) 2] * [cosx/(1+cosx)]

=[sin2x*(cos2x)^2*cosx]/[2(cos2x)^2*(cosx)^2*( 1+cosx)]

=[sin2x*cosx]/2(cosx)^2*( 1+cosx)]

=[2sinx*(cosx)^2/[(cosx)^2*( 1+cosx)]

=sinx/( 1+cosx)

=[2sinx/2*cosx/2]/[2(cosx/2)^2]

=(sinx/2)/(cosx/2)

=tan(x/2)

Left = right

Seven: A+B= π/4

B= π/4-A, because Tan (π/4-a) = (1-tana)/(1+tana).

( 1+tanA)( 1+tanB)=( 1+tanA)[ 1+tan(π/4-A)]

=( 1+ tana) {1+[( 1- tana) /( 1+ tana)]}

=( 1+ tana) +( 1- tana)

=2

Therefore, (1+tana) (1+tanb) = 2.