Question

If cosec θ - sin θ=a and sec θ - cos θ=b , then find the value of (a2b)23+(ab2)23.

Answer 1

Step-by-step explanation:

Given : cosecθ−sinθ=a3 and secθ−cosθ=b3

⇒sinθ1−sinθ=a3

⇒sinθcos2θ=a3

⇒a=(sinθcos2θ)31

Similarly ; b=(cosθsin2θ)31

⇒a2b2(a2+b2)=a4b2+a2b4

⇒(sinθcos2θ)34(cosθsin2θ)

please mark me as a brilliant please mark me

Answer 2

Answer:

Step-by-step explanation:

Given : cosecθ−sinθ=a3 and secθ−cosθ=b3

⇒sinθ1−sinθ=a3

⇒sinθcos2θ=a3

⇒a=(sinθcos2θ)31

Similarly ; b=(cosθsin2θ)31

⇒a2b2(a2+b2)=a4b2+a2b4

⇒(sinθcos2θ)34(cosθsin2θ)