If cosec-sin =m^3 and sec -cos =n^3 then prove that m^4n^2+m^2n^4=1.
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m⁴n² + m²n⁴ = 1 if Cosec - Sin = m³ & sec -cos =n³
Step-by-step explanation:
Cosec - Sin = m³
=> 1/Sin - Sin = m³
=> 1 - Sin² = m³Sin
=> Cos² = m³Sin
=> m³ = Cos²/Sin
sec -cos =n³
=1/Cos - Cos = n³
=> Sin² = n³Cos
=> Sin²/Cos² = n³Cos/m³Sin
=> Sin³/Cos³ = n³/m³
=> Sin/Cos = n/m
=> tan = n/m
=> n = mtan
LHS
= m⁴n² + m²n⁴
= m⁴(mtan)² + m²(mtan)⁴
= m⁶tan² + m⁶tan⁴
= (m³)²tan²(1 + tan²)
= (Cos²/Sin)² tan²Sec²
=Cos² * Cos² * tan²Sec²/Sin²
= Cos² * Sin² * sec²/Sin²
= Cos² * Sec²
= 1
= RHS
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