Question

Cosec -sin=a3 and sec-cos =b3 prove that a2b2 (a2+b2)=1

Answer 1

:  1/sinθ - sinθ = a^3  

(1 - sin²θ)/sinθ = a^3  

cos²θ / sinθ = a^3  

similarly  

sin²θ / cosθ = b^3  

then  

a^4 b^2 + a^2 b^4  

= [cos²θ/sinθ]^(4/3) [sin²θ / cosθ]^(2/3) + [cos²θ/sinθ]^(2/3) [sin²θ / cosθ]^(4/3)  

= (cosθ)^(8/3) / (cosθ)^(2/3) + (sinθ)^(8/3) / (sinθ)^(2/3)  

= cos²θ + sin²θ  

= 1  

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