Question

If sinϴ + cosecϴ = 2, then find the value of sin2ϴ + cosec2ϴ

Answer 1

Answer:

We have,

(sin θ + cosec θ) = 2

squaring both sides, we get    

(sin θ + cosec θ)^2 = 2^2 ,{(a+b)^2=a^2+b^2+2ab)}

sin2θ + cosec2θ + 2×sin θ×cosec θ = 4

sin2θ + cosec2θ + 2×sin θ×1sin θ = 4

sin2θ + cosec2θ + 2 = 4

sin2θ + cosec2θ = 4 − 2

sin2θ + cosec2θ = 2

Answer 2

Answer:

sin2θ.cosec2θ  

=sin2θ×1sin2θ                  [∴cosecθ=1sinθ]

=1