Question

cos (90-tetha)/1+sin (90-tetha)+1+sin(90-tetha)/cos (90-tetha)=2cosec tetha

Answer 1

Step-by-step explanation:

L.H.S

={cos(90-θ)}/{1+sin(90-θ)}+{1+sin(90-θ)}/{cos(90-θ)}

={sinθ/1+cosθ}+{1+cosθ/sinθ}

={sin²θ+(1+cos²θ)}/sinθ(1+cosθ)}

=sin²θ+1+2cosθ+cos²θ/sinθ(1+cos)

=sin²θ+cos²θ+1+2cosθ/sinθ(1+sinθ)

=2+2cosθ/sinθ(1+cosθ)

=2(1+cos)/sinθ(1+cosθ)

=2/sinθ

=2 cosecθ

proved

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