English

`Cot Theta/((Cosec Theta + 1) )+ ((Cosec Theta +1 ))/ Cot Theta = 2 Sec Theta `

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Question

`cot theta/((cosec  theta + 1) )+ ((cosec  theta +1 ))/ cot theta = 2 sec theta `

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Solution

LHS = `cot theta/((cosec  theta + 1) )+ ((cosec  theta +1 ))/ cot theta `

      =`( cot ^2 theta + (cosec  theta + 1 ) ^2 ) / ((cosec  theta +1) cot theta)`

      =` ( cot ^2 + cosec ^2 theta + 2 cosec  theta +1 )/( (cosec  theta +1) cot theta)`

      =`( cot ^2  theta + cosec ^2  theta +2cosec  theta + cosec ^2  theta - cot^2 theta)/((cosec theta +1 ) cot theta)`

      =` (2 cosec^2  theta + 2 cosec  theta)/(( cosec  theta +1 ) cot theta)` 

      =`(2 cosec  theta ( cosec  theta +1))/(( cosec  theta +1 ) cot theta)`

      =` (2 cosec  theta)/(cot theta)`

      =`2 xx 1/sin  theta xx sin theta/ cos theta`

      = 2 sec 𝜃
       = RHS
Hence, LHS = RHS

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Chapter 13: Trigonometric identities - Exercises 1

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 13 Trigonometric identities
Exercises 1 | Q 19.2
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