Question

if sin alpha+cosec alpha=2,find the value of sin*nalpha+cosec*nalpha, n belongs to intezers​

Answer 1

Step-by-step explanation:

follow the above steps

Answer 2

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

$sin \alpha + cos\alpha = 2$

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

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,

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

Hence the value of sin a is 1 .

So , now ;

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

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

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$\large\underline{\purple{\bf Extra\: Information:- }}$

➳ $\sf\red{ cosec^2\theta - cot^2\theta=1}$

➳ $\sf\red{ sin^2\theta + cos^2\theta=1}$

➳ $\sf\red{ sec^2\theta - tan^2\theta=1}$

➳ $\sf\red{ sin(A+B)=sinA.cosB+cosA.sinB}$

➳ $\sf\red{ sin(A-B)=sinA.cosB-cosA.sinB}$

➳ $\sf\red{cos(A+B)=cosA.cosB-sinA.sinB }$

➳ $\sf\red{cos(A-B)=cosA.cosB+sinA.sinB }$

➳ $\sf\red{ tan(A+B)=\dfrac{tanA+tanB}{1-tanA.tanB}}$

➳ $\sf\red{ tan(A-B)=\dfrac{tanA-tanB}{1+tanA.tanB}}$

➳ $\sf\red{ tan(A+B+C)=\dfrac{tanA+tanB+tanC-tanA.tanB.tanC}{1-tanA.tanB-tanB.tanC-tanC.tanA}}$

$\rule{200}2$

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$