Math, asked by doresh, 1 year ago

sin alpha+cosec alpha=2

Answers

Answered by RabbitPanda
6
Let alpha be x.
Then If your question is sinnx + cosecnx then the answer is as follows:
sinx + 1/sinx = 2
[sin2x + 1]/sinx=2
sin2x + 1 =2sinx
sin2x - 2sinx + 1 = 0
(sinx-1) (sinx-1) = 0
So sinx=1
Since sinx =1 , 1/sinx =1 i.e. cosecx =1
So sinnx + cosecnx = 1n + 1n = 2

@skb
Answered by Anonymous
0

\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}}}

__________________________________

\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}}

Similar questions