Math, asked by viratkohli30228, 7 months ago

prove that LHS=RHS.
ans the above question​

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Answered by Anonymous
3

Question:-

 :  \implies \sqrt{ \dfrac{1 +  \cos\theta}{1 -  \cos\theta } }  =  \csc \theta  +  \cot\theta

Solution:-

 :  \implies \sqrt{ \dfrac{1 +  \cos\theta}{1 -  \cos\theta } }  =  \csc \theta  +  \cot\theta

Now

   \to \sqrt{ \dfrac{1 +  \cos\theta}{1 -  \cos\theta } }  =   \dfrac{1}{ \sin \theta }   +   \dfrac{ \cos \theta }{ \sin\theta}

\to \sqrt{ \dfrac{1 +  \cos\theta}{1 -  \cos\theta } }  =   \dfrac{1 +  \cos \theta }{ \sin \theta  }

 \to \dfrac{1 +  \cos \theta }{1 -  \cos \theta }  =  \dfrac{(1 +  \cos \theta)  {}^{2} }{ \sin {}^{2}  \theta }

Apply this identity in RHS

=> ( a + b ) ²= a² + b² + 2ab

we get

 \rm\to \dfrac{1 +  \cos \theta }{1 -  \cos \theta }  =  \dfrac{(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }{ \sin {}^{2}  \theta }

\rm\to \dfrac{1 +  \cos \theta }{1 -  \cos \theta }  =  \dfrac{(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }{( 1 -  \cos {}^{2} \theta) }

Using this identities

( a² - b² ) = ( a - b ) ( a + b )

\rm\to \dfrac{1 +  \cos \theta }{1 -  \cos \theta }  =  \dfrac{(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }{( 1 -  \cos {}^{} \theta) (1 +  \cos \theta) }

\rm\to \dfrac{1 +  \cos \theta }{ \cancel{1 -  \cos \theta} }  =  \dfrac{(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }{ \cancel{( 1 -  \cos {}^{} \theta)} (1 +  \cos \theta) }

\rm\to {1 +  \cos \theta }{  }  =  \dfrac{(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }{ {} 1 +  \cos \theta}

\rm\to {(1 +  \cos \theta)(1 +  \cos \theta)  }{  }  =  {(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }

\rm\to {(1 +  \cos \theta) {}^{2}   }{  }  =  {(1 +  \cos {}^{2}  \theta + 2 \cos\theta)  {}^{} }

 \rm \to{1 +  \cos {}^{2}  \theta + 2 \cos\theta {}^{} }  = {1 +  \cos {}^{2}  \theta + 2 \cos\theta {}^{} }

Hence proved

Some trigonometry identities which we use in this question

 \rm \:  \implies \csc \theta =  \dfrac{1}{ \sin \theta }

 \rm \implies \:  \cot( \theta)  =  \dfrac{ \cos( \theta) }{ \sin( \theta) }

 \rm \:  \implies \: 1 -  \cos {}^{2} ( \theta)  =  \sin {}^{2} ( \theta)

Answered by Lueenu22
0

Step-by-step explanation:

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