if cosec 0 - sin 0=a sec 0 -cos 0=b,prove that a²b²(a²+b²+3)=1
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Answer: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
Step-by-step explanation: