Question

Sin theta/ 1-cos theta+tan theta/1+cos theta = cot theta+sec theta×cosec theta

Answer 1

Question:

To Prove :

${\sf{ {\dfrac{sin \theta}{1 - cos \theta}} + {\dfrac{tan \theta}{1 + cos \theta}} = cot \theta + sec \theta cosec \theta}}$

Step-by-step explanation:

L.H.S. = ${\sf{ {\dfrac{sin \theta}{1 - cos \theta}} + {\dfrac{tan \theta}{1 + cos \theta}}}}$

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$\implies{\sf{ {\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{(1 + cos \theta)(1 - cos \theta)}}}}$

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${\boxed{\tt{Identity \ : \ (a + b)(a - b) = a^2 - b^2}}}$

${\tt{Here, \ a = 1, \ b = cos \theta}}$

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$\implies{\sf{ {\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{(1)^2 - (cos \theta)^2}}}}$

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$\implies{\sf{ {\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{1 - cos^2 \theta}}}}$

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${\boxed{\tt{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}$

${\tt{From \ this, \ we \ get \ [ 1 - cos^2 \theta = sin^2 \theta ]}}$

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$\implies{\sf{ {\dfrac{ sin \theta (1 + cos \theta ) + tan \theta (1 - cos \theta )}{sin^2 \theta}}}}$

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$\implies{\sf{ {\dfrac{ sin \theta + sin \theta cos \theta + tan \theta - tan \theta cos \theta}{sin^2 \theta}}}}$

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${\boxed{\tt{Identity \ : \ tan \theta = {\dfrac{sin \theta}{cos \theta}}}}}$

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$\implies{\sf{ {\dfrac{ sin \theta + sin \theta cos \theta + tan \theta - {\dfrac{sin \theta}{cos \theta}} \times cos \theta}{sin^2 \theta}}}}$

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$\implies{\sf{ {\dfrac{ sin \theta + sin \theta cos \theta + tan \theta - sin \theta}{sin^2 \theta}}}}$

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Rearranging the terms in the numerator.

$\implies{\sf{ {\dfrac{ sin \theta - sin \theta + sin \theta cos \theta + tan \theta}{sin^2 \theta}}}}$

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$\implies{\sf{ {\dfrac{sin \theta cos \theta + tan \theta}{sin^2 \theta}}}}$

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${\boxed{\tt{Identity \ : \ tan \theta = {\dfrac{sin \theta}{cos \theta}}}}}$

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$\implies{\sf{ {\dfrac{sin \theta cos \theta + {\dfrac{sin \theta}{cos \theta}}}{sin^2 \theta}}}}$

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$\implies{\sf{ {\dfrac{ {\dfrac{sin \theta cos \theta (cos \theta) + sin \theta}{cos \theta}} }{ sin^2 \theta}} }}$

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$\implies{\sf{ {\dfrac{ {\dfrac{sin \theta cos^2 \theta + sin \theta}{cos \theta}} }{ sin^2 \theta}} }}$

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$\implies{\sf{ {\dfrac{sin \theta cos^2 \theta + sin \theta }{ (cos \theta)(sin^2 \theta)}}}}$

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$\implies{\sf{ {\dfrac{sin \theta cos^2 \theta + sin \theta }{cos \theta . sin^2 \theta}}}}$

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Taking common terms out from numerator and denominator.

$\implies{\sf{ {\dfrac{sin \theta (cos^2 \theta + 1) }{sin \theta (cos \theta . sin \theta)}}}}$

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$\implies{\sf{ {\dfrac{cos^2 \theta + 1}{cos \theta . sin \theta}}}}$

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We can write this as :

$\implies{\sf{ {\dfrac{cos^2 \theta}{cos \theta sin \theta}} + {\dfrac{1}{cos \theta sin \theta}} }}$

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$\implies{\sf{ {\dfrac{cos \theta}{sin \theta}} + {\dfrac{1}{cos \theta sin \theta}} }}$

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We can write this as :

$\implies{\sf{ {\dfrac{cos \theta}{sin \theta}} + {\dfrac{1}{cos \theta}} \times {\dfrac{1}{sin \theta}} }}$

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${\boxed{\tt{Identity \ : \ {\dfrac{cos \theta}{sin \theta}} = cot \theta}}}$

${\boxed{\tt{Identity \ : \ {\dfrac{1}{cos \theta}} = sec \theta}}}$

${\boxed{\tt{Identity \ : \ {\dfrac{1}{sin \theta}} = cosec \theta}}}$

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$\implies{\sf{ cot \theta + sec \theta cosec \theta}}$

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= R.H.S.

Hence, proved !!