Question

If cosecθ - sinθ = a3, secθ - cosθ = b3, prove that a²b²(a² + b²).

Answer 1

Given:

cosecθ - sinθ = a³   -(1)

secθ - cosθ = b³.     -(2)

To find:

Solution to a²b²(a² + b²)

Answer:

• By solving these equations by rewriting cosec as inverse of sin and sec as inverse of cos, we get values for a^2 and b^2.
• Equation 1,
• $\frac{1}{sin x} - sin x =a^{3}$
• $\frac{1 - sin^{2} x}{sin x} =a^{3}$
• $(\frac{cos^{2} x}{sin x})^{2/3} =(a^{3} )^{2/3}$
• $\frac{cos^{4/3} x}{sin ^{2/3} x} = a^{2}$   -(3)
• Equation 2,
• secθ - cosθ = b³
• $\frac{1}{cos x} - cos x = b^{3}$
• $(\frac{1 - cos^{2} x}{cos x} )^{2/3} = (b^{3})^{2/3}$
• $\frac{sin ^{4/3} x}{cos ^{2/3} x} = b^{2}$    -(4)
• Multiplying 3 and 4,
• $a^{2} b^{2} = sin^{2/3} x .cos^{2/3} x$   -(5)
• Adding 3 and 4,
• We also get values of a²+ b²
• $a^{2} + b^{2} = \frac{cos^{4/3} x}{sin^{2/3} x} + \frac{sin ^{4/3} x}{cos^{2/3} x}$      -(6)
• By multiplying  5 and 6 we get
• $a^{2}b^{2} (a^{2} +b^{2} ) = \frac{1}{sin ^{2/3} x cos ^{2/3} x} * sin^{2/3} x * cos^{2/3} x$

a²b²(a² + b²)= 1

Answer 2

Answer:

Step-by-step explanation:

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