Question

If \(\operatorname{cosec} \theta-\sin \theta=m \) and \(\sec \theta-\cos \theta=\mathbf{n}\) then what is \(m^{\frac{4}{3}} n^{\frac{2}{3}}+m^{\frac{2}{3}} n^{\frac{4}{3}}\) equal to ?

Answer 1

Correct Answer - Option 2 : 1

Given : 

cosec θ - sin θ = m 

sec θ - cos θ = n 

Calculations :

Put θ = 45° 

cosec 45°  - sin 45° = m 

⇒ m = √2 - (1/√2) 

⇒ m = 1/√2 

sec 45° - cos 45° = n 

⇒ n = √2 - (1/√2) 

⇒ n = 1/√2   

Putting value of n and m in the equation 

m^4/3 . n^2/3 + m^2/3 . n^4/3 = (1/√2)^4/3 . (1/√2)^2/3 + (1/√2)^2/3 . (1/√2)^4/3

⇒ (1/√2)² + (1/√2)² 

⇒ (1/2) + (1/2) 

⇒ 1 

∴ Option 2 will be the correct answer.