Question

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

Answer 1

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