Question

if cosectheta-sintheta=a,sectheta-costheta=b,Prove thata^2b^2(a^2+b^2+3)=1​

Answer 1

Answer:

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Answer 2

Answer with Step-by-step explanation:

$cosec\theta-sin\theta=a$

$\frac{1}{sin\theta}-sin\theta=a$

$\frac{1-sin^2\theta}{sin\theta}=a$

$\frac{cos^2\theta}{sin\theta}=a$

Using the formula

$cos^2\theta=1-sin^2\theta$

$sec\theta-cos\theta=b\\\frac{1}{cos\theta}-cos\theta=b\\sec\theta=\frac{1}{cos\theta}\\\frac{1-cos^2\theta}{cos\theta}=b\\\frac{sin^2\theta}{cos\theta}=b$

RHS

$(sin^2\theta+cos^2\theta)^3=(cos^2\theta)^3+(sin^2\theta)^3+3sin^2\thetacos^2\theta(sin^2\theta+cos^2\theta)$

By using identity

$(a+b)^3=a^3+b^3+3ab(a+b)$

$1=(cos^2\theta)^3+(sin^2\theta)^3+3 sin^2\theta cos^2\theta$

Using the formula

$sin^2\theta+cos^2\theta=1$

LHS

$a^2b^2(a^2+b^2+3)$

Substitute the  values

$\frac{(cos^2\theta)^2}{sin^2\theta}\times\frac{(sin^2\theta)^2}{cos^2\theta}(\frac{(cos^2\theta)^2}{sin^2\theta}+\frac{(sin^2\theta)^2}{cos^2\theta}+3)$

$sin^2\theta cos^2\theta(\frac{(cos^2\theta)^2}{sin^2\theta}+\frac{(sin^2\theta)^2}{cos^2\theta}+3)$

$(cos^2\theta)^3+(sin^2\theta)^3+3sin^2\theta cos^2\theta$

LHS=RHS

Hence, proved

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