Question

3. It " Sin aloha+cosec alpha=2, find
the value of sin n alpha+cosec n aloha n belong to t​

Answer 1

$\large\underline{\purple{\sf Given :- }}$

• $sin \alpha + cos\alpha = 2$

$\large\underline{\purple{\sf To\:\:Find :- }}$

• The value of $sin^n \alpha + cosec^n\alpha$

$\large\underline{\purple{\sf Answer :- }}$

It's given that sin a + csc a = 2 . And we are required to find the value of sin ⁿ a + cosⁿ a . So,

$=> sin\alpha + cosec\alpha = 2 \\=> sin\alpha +\frac{1}{sin\alpha}=2 \\=> \dfrac{sin^2\alpha+1}{sin\alpha}=2 \\=> sin^2\alpha + 1 = 2sin\alpha \\=> sin^2\alpha-2sin\alpha+1=0\\=> sin^2\alpha-sin\alpha-sin\alpha+1 = 0 \\=> sin\alpha (sin\alpha -1)-1(sin\alpha-1)=0\\=> (sin\alpha-1)(sin\alpha-1)=0\\=> (sin\alpha-1)^2 \\=> \bf sin\alpha = 1,1$

Hence the value of sin a is 1 .

So , now ;

$=> sin^n\alpha + cosec^n\alpha = 1^n + 1^n \\=> sin^n \alpha + cosec^n\alpha = 1+1 \\=>\bf sin^n\alpha + cosec^n\alpha = 2$

$\Large{\boxed{\pink{\sf \purple{\bigstar}\:sin^n\alpha+cos^n\alpha = 2}}}$

Answer 2

Answer:

sinα+cosecα=2

sinα+

sinα

1

=2

sin

2

α+1=2sinα

sin

2

α−2sinα+1=0

(sinα−1)

2

=0

sinα=1

hence,

cosecα=

sinα

1

=1

now,

sin

n

α+csc

n

α=(1)

n

+(1)

n

=1+1=2