Solution: left = (tanx/2+1)/(kloc-0/-tanx/2)+(tanx/2-1)/(1+tanx/2)

=(4tanx/2)/( 1-tan？ x/2)

=2tanx= right

Question 8:

Solution: left = (1-cos2θ+sin2θ)/(1+cos2θ+sin2θ)

=(2sin？ θ+2sinθcosθ)/(2cos？ θ+2sinθcosθ)

= {sin θ (sin θ+cos θ)}/{cos θ (sin θ+cos θ)} = right.