prove: (sinθ + tanθ)/(cosθ + 1) = tanθ?

..sinθ + tanθ

-------------------..... note tanθ = sinθ/cosθ

..cosθ + 1

....sinθ + sinθ/cosθ.......cosθ

=------------------------- x ----------

.....cosθ + 1...................cosθ

..sinθcosθ + sinθ

=------------------------

.....cosθ(cosθ + 1)

...sinθ(cosθ + 1)

=----------------------- cancel out cosθ + 1

....cosθ(cosθ + 1)

= tanθ prove//

tanθ = sinθ/cosθ so,

(sinθ + (sinθ/cosθ))/(cosθ + 1)

=> [(sinθcosθ + sinθ)/cosθ]/(cosθ + 1)

i.e. [sinθ(cosθ + 1)/cosθ]/(cosθ + 1)

=> [sinθ/cosθ](cosθ + 1)/(cosθ + 1)

so, sinθ/cosθ => tanθ 

:)>

To prove that-

(sinθ + tanθ)/(cosθ + 1) = tanθ

...........................sin θ

LHS =   [ sin θ + ---------  ] divided by ( cos θ + 1 )

.......................... cos θ

...... sin θ * cos θ + sin θ

=> --------------------------------  divided by ( 1 + cos θ )

................... cos θ

.... sin θ ( 1 + cos θ )

=>--------------------------  divided by ( 1 + cos θ )

............ cos θ

.......sin θ ( 1 + cos θ )...............1.. 

=> --------------------------- x -----------------   

............... cos θ ..................( 1 + cos θ ). 

............ sin θ

=>---------------   =  tan θ =  RHS  .................... Proved

.......... cos θ