Question

if cosec theta - sin theta= a^3, sec theta-cos theta= b^3, then the value of a^2b^2(a^2+b^2) is. (a)2. ( b)3. (c)4. (d)1​

Answer 1

(Instead of θ, I use A)

Answer:

Option (d) 1

Step-by-step explanation:

Given :

cosec A - sin A = a³,  ...(i)

sec A - cos A = b³,   ...(ii)

Then, To find the value of,

a²b²(a² + b²),.

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Solution :

We know that,

$cosec A = \frac{1}{sin A}$ ...(iii)

$sec A = \frac{1}{cos A}$    ...(iv)

$sin^2 A + cos^2A = 1$      ...(v)

Hence,

(i) ⇒ $cosec A - sin A = a^3$

⇒ $\frac{1}{sin A}- sin A = a^3,$

⇒ $\frac{1-sin^2A}{sin A} = a^3,$  from (v)

⇒ $\frac{cos^2A}{sin A} = a^3,$

⇒ $(\frac{cos^2A}{sinA})^{\frac{2}{3}} = a^{3(\frac{2}{3})}$

⇒ $\sqrt[3]{(\frac{cos^2A}{sinA})^2} = a^2$

⇒ $\sqrt[3]{\frac{cos^4A}{sin^2A}}= a^2$...(vi)

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(ii) ⇒ $sec A - cos A = b^3$

⇒ $\frac{1}{cos A}  - cos A = b^3,$

⇒ $\frac{1-cos^2A}{cos A} = b^3,$  from (v)

⇒ $(\frac{sin^2A}{cosA})^{\frac{2}{3} } = b^{3(\frac{2}{3})}$

⇒ $\sqrt[3]{(\frac{sin^2A}{cosA})^2} = b^2$

⇒ $\sqrt[3]{\frac{sin^4A}{cos^2A}}= b^2$...(vi)

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Hence,

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

⇒ $(\sqrt[3]{\frac{cos^4A}{sin^2A}})(\sqrt[3]{\frac{sin^4A}{cos^2A}})(\sqrt[3]{\frac{cos^4A}{sin^2A}}+\sqrt[3]{\frac{sin^4A}{cos^2A}})$

⇒ $\sqrt[3]{\frac{cos^4Asin^4A}{sin^2Acos^2A}}(\frac{\sqrt[3]{cos^4A} }{\sqrt[3]{sin^2A}} +\frac{\sqrt[3]{sin^4A} }{\sqrt[3]{cos^2A} } )$

⇒ $(\sqrt[3]{sin^2Acos^2A})(\frac{(\sqrt[3]{cos^4A})(\sqrt[3]{cos^2A})+(\sqrt[3]{sin^4A})(\sqrt[3]{sin^2A}}{(\sqrt[3]{sin^2A})(\sqrt[3]{cos^2A})}$

⇒ $(\sqrt[3]{sin^2Acos^2A})(\frac{(\sqrt[3]{cos^6A})+(\sqrt[3]{sin^6A})}{\sqrt[3]{sin^2Acos^2A}}$

⇒ $\sqrt[3]{(cos^2A)^3} + \sqrt[3]{(sin^2A)^3}= cos^2A + sin^2A = 1$