Question

please answer this question

Answer 1

=(1+tanA+secA)(1+cotA-cosecA)

=(1+sinA/cosA+1/cosA)(1+cosA/sinA-1/sinA)

={(cosA+sinA+1)/cosA}{(sinA+cosA-1)/sinA}

={(cosA+sinA)²-1²}/cosAsinA

=(cos²A+sin²A+2sinAcosA-1)/cosAsinA

=(1+2sinAcosA-1)/cosAsinA

=2cosAsinA/cosAsinA

=2

Answer 2

(1+tan θ+sec θ)(1+cot θ-cosec θ)
=1+tanθ+secθ+cotθ+tanθ×cotθ+cotθ×secθ-cosecθ-cosecθ×tanθ-cosecθsec =1+tanθ+secθ+cotθ+1+(cosθ/sinθ)(1/cosθ)-cosecθ-(1/sinθ)(sinθ/cosθ)             -(1/sinθ)(1/cosθ)
=2+(sinθ/cosθ)+(1/cosθ)+(cosθ/sinθ)+(1/sinθ)-(1/sinθ)-(1/cosθ)-(1/sinθ×cosθ)
=2+[(sin²θ+sinθ+cos²θ-sinθ-1)/(sinθ×cosθ)]
=2+[0/(sin θ×cos θ)]
=2

Hope it helps u :-)