Question

1+cosx÷1-cosx+1-cosx÷1+cosx=4cotxcosecx

Answer 1

=(1+cosx)/(1-cosx)-(1-cosx)/1+cosx)
={(1+cosx)^2-(1-cosx)^2}/(1-cos^2x)
={1+cos^2x+2cosx-(1+cos^2x-2cosx)}/sin^2x
=(1+cos^2x+2cosx-1-cos^2x+2cosx)/sin^2x
=4cosx/sin^2x
=4cosx/sinx×sinx
=4×cosx/sinx×1/sinx
=4cotxcosecx

Answer 2

$\frac{1 + cosx}{1 - cosx} - \frac{1 - cosx}{1 + cosx} \\ \\ = \frac{( {1 + cosx)}^{2} - ( {1 - cosx)}^{2} }{1 - {cos}^{2}x } \\ \\ = \frac{4.1.cosx}{ {sin}^{2}x } \\ \\ = 4 \frac{cosx}{ {sin}^{2} x} \\ \\ = 4cotxcosecx$