Question

if cosecA- sinA=a^3 and secA-cosA=b^3 the find a^2b^2 (a^2+b^2)​

Answer 1

Question

if cosec A- sin A=a³ and sec A-cos A=b³ the find a²b² (a²+b²)

Solution

Given :-

• cosec A- sin A=a³ --------(1)
• sec A - cos A = b³ ---------(2)

Find :-

• a²b²(a² + b²) = ?

Explanation

Important Formula

★ Sin² x + Cos² x = 1

★ Sin x = 1/(Cosec x)

★ Sec x = 1/(Cos x)

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So, by equ(1)

➠ 1/sin A - sin A = a³

➠ (1 - sin² A)/sin A = a³

Then,

➠ cos² A / sin A = a³

So, Value of a will be

➠ a² = [ cos² A / sin A]^(2/3) --------(3)

Now, by equ(2)

➠1/cos A - cos A = b³

➠ ( 1- cos² A)/cos A = b³

➠ sin² A / cos A = b³

So, Value of b will be

➠ b² = [ sin² A / cos A] ^(2/3) --------(4)

Now, Calculate Value of a²b²

➠ a²b²

keep Value of a² & b²

➠a²b² = [ cos² A / sin A]^(2/3) * [ sin² A / cos A] ^(2/3)

➠a²b² = (cos A )^(2 /3) * (sin A )^(2/3)

Now, Calculate Value of (a²+b²)

➠ a²+b²

Keep Value of a² & b²

➠ (a²+b²) = [ cos² A / sin A]^(2/3) + [ sin² A / cos A] ^(2/3)

➠ (a²+b²) = [(cos³ A + sin³ A)/(sin A. cos A) ]^(2/3)

Now,

➠ a²b² (a²+b²) = (cos A )^(2 /3) * (sin A )^(2/3) * [(cos³ A + sin³ A)/(sin A. cos A) ]^(2/3)

➠ a²b² (a²+b²) = [(cos³ A + sin³ A)]^(2/3)

➠ a²b² (a²+b²) = (cos A )^(3 × 2/3) + (sin A)^(3 × 2/3)

➠ a²b² (a²+b²) = cos² A + sin² A

[ ★ ( cos² A + sin² A) = 1 ]

➠ a²b² (a²+b²) = 1

___________________________

Hence

• Value of a²b² (a²+b²) = 1

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Answer 2

★ Answer refer to the attachment ★