Question

Given sec A = 2 Find the value of
(sin^3 A + cos^3 A)/(sin A + cos A) +
(sin^3 A - cos^3 A)/(sin^2 A - cos A)​

Answer 1

Step-by-step explanation:

Given :-

Sec A = 2

To find :-

Value of [(sin³ A + cos³ A)/(sin A + cos A )] + [(sin³ A -cos³ A) /(sin² A - cos A)]

Solution :-

Given that sec A = 2

=> 1 / cos A = 2

=> cos A = 1/2 -----------(1)

We know that

sin² A + cos² A = 1

=> sin² A + (1/2)² = 1

=> sin² A + (1/4) = 1

=> sin² A = 1-(1/4)

=> sin² A = (4-1)/4

=> sin² A = 3/4

=> sin A = √(3/4)

=> sin A = √3/2 ---------(2)

now,

(sin³ A + cos³ A)/(sin A+ cos A)

= (sin A+cos A)(sin² A+cos² A-sinAcos A) / (sin A+ cos A)

Since, a³+b³ = (a+b)(a²-ab+b²)

=>(sin A+cos A)(1-sin Acos A)/(sin A+cos A)

=> 1-sin A cos A

=> 1-[(√3/2)(1/2)]

=> 1-(√3/4)

=> (4-√3)/4

(sin³ A+cos³ A)/(sin A+cos A) = (4-√3)/4

and

sin³ A - cos³ A

=>(sin A-cos A)(sin² A+cos² A+sinAcos A)

since, a³-b³ = (a-b)(a²+ab+b²)

=> (sin A - cos A)(1+sin A cos A)

=> [(√3/2)-(1/2)][1+(√3/2)(1/2)]

=> [(√3-1)/2][1+(√3/4)]

=> [(√3-1)/2][(4+√3)/4]

=> (√3-1)(4+√3)/8

=> (4√3+3-4-√3)/8

=> (3√3-1)/8

sin³ A - cos³ A = (3√3-1)/8

and

sin² A-cos A

=> (√3/2)²-(1/2)

=> (3/4)-(1/2)

=> (3-2)/4

=> 1/4

sin² A-cos A = 1/4

Now,

(sin³ A - cos³ A)/(sin² A-cos A)

=> [(3√3-1)/8]/(1/4)

=> [(3√3-1)/8]×(4/1)

=> [4(3√3-1)/8]

=> (3√3-1)/2

(sin³ A-cos³ A)/(sin² A-cos A) = (3√3-1)/2

now,

[(sin³ A + cos³ A)/(sin A + cos A )] +

[(sin³ A -cos³ A) /(sin² A - cos A) ]

=> [(4-√3)/4] + [ (3√3-1)/2]

=> [(4-√3)+2(3√3-1)]/4

=> (4-√3+6√3-2)/4

=> (2+5√3)/4

Answer :-

[(sin³ A + cos³ A)/(sin A + cos A ) ]+

[(sin³ A -cos³ A) /(sin² A - cos A)] = (2+5√3)/4

Used formulae:-

→ sin² A + cos² A = 1

→ sec A = 1 / cos A

→ a³+b³ = (a+b)(a²-ab+b²)

→ a³-b³ = (a-b)(a²+ab+b²)

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given :-

Sec A = 2

To find :-

Value of [(sin³ A + cos³ A)/(sin A + cos A )] + [(sin³ A -cos³ A) /(sin² A - cos A)]

Solution :-

Given that sec A = 2

=> 1 / cos A = 2

=> cos A = 1/2 -----------(1)

We know that

sin² A + cos² A = 1

=> sin² A + (1/2)² = 1

=> sin² A + (1/4) = 1

=> sin² A = 1-(1/4)

=> sin² A = (4-1)/4

=> sin² A = 3/4

=> sin A = √(3/4)

=> sin A = √3/2 ---------(2)

now,

(sin³ A + cos³ A)/(sin A+ cos A)

= (sin A+cos A)(sin² A+cos² A-sinAcos A) / (sin A+ cos A)

Since, a³+b³ = (a+b)(a²-ab+b²)

=>(sin A+cos A)(1-sin Acos A)/(sin A+cos A)

=> 1-sin A cos A

=> 1-[(√3/2)(1/2)]

=> 1-(√3/4)

=> (4-√3)/4

(sin³ A+cos³ A)/(sin A+cos A) = (4-√3)/4

and

sin³ A - cos³ A

=>(sin A-cos A)(sin² A+cos² A+sinAcos A)

since, a³-b³ = (a-b)(a²+ab+b²)

=> (sin A - cos A)(1+sin A cos A)

=> [(√3/2)-(1/2)][1+(√3/2)(1/2)]

=> [(√3-1)/2][1+(√3/4)]

=> [(√3-1)/2][(4+√3)/4]

=> (√3-1)(4+√3)/8

=> (4√3+3-4-√3)/8

=> (3√3-1)/8

sin³ A - cos³ A = (3√3-1)/8

and

sin² A-cos A

=> (√3/2)²-(1/2)

=> (3/4)-(1/2)

=> (3-2)/4

=> 1/4

sin² A-cos A = 1/4

Now,

(sin³ A - cos³ A)/(sin² A-cos A)

=> [(3√3-1)/8]/(1/4)

=> [(3√3-1)/8]×(4/1)

=> [4(3√3-1)/8]

=> (3√3-1)/2

(sin³ A-cos³ A)/(sin² A-cos A) = (3√3-1)/2

now,

[(sin³ A + cos³ A)/(sin A + cos A )] +

[(sin³ A -cos³ A) /(sin² A - cos A) ]

=> [(4-√3)/4] + [ (3√3-1)/2]

=> [(4-√3)+2(3√3-1)]/4

=> (4-√3+6√3-2)/4

=> (2+5√3)/4

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