Question

 If k = Cosec + Cot then prove that k^2 - 1 / k^2 + 1 = cos

Answer 1

HELLO DEAR,

K = COSEC∅ + COT∅-------------(1)

we know that:-

cosec²∅ - cot²∅ = 1

(cosec∅ - cot∅)(cosec∅ + cot∅) = 1

from--(1)

(cosec∅ - cot∅) × k = 1

=> (cosec∅ - cot∅) = k-----------(2)


now adding---(1) and---(2)

we get,


COSEC∅ + COT∅ + cosec∅ - cot∅ = k + 1/k

=> 2cosec∅ = (k² + 1)/k

=> 2/sin∅ = (k²+1)/k

=> sin∅ = 2k/(k²+1)

on squaring Both side,

we get,

sin²∅ = 4k²/(k²+1)²

=> ( 1 - cos²∅) = 4k²/(k²+1)²

=> cos²∅ = 1 - 4k²/(k²+1)²

=> cos²∅ =
$\sqrt{ \frac{ {k}^{4} + 1 + 2 {k}^{2} - 4 {k}^{2} }{ ({ {k}^{2} + 1})^{2} } } \\ \\ = > \sqrt{ \frac{ {k}^{4} + 1 - 2 {k}^{2} }{ {( {k}^{2} + 1) }^{2} } } \\ \\ = > \sqrt{ \frac{ {( {k}^{2} - 1) }^{2} }{ {( {k}^{2} + 1) }^{2} } } = \frac{ {( {k}^{2} - 1) } }{ {( {k}^{2} + 1) } }$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

HEYA!!!!

Refer the attachment for your answer.

HOPE THIS HELPS U...