If cosec A = x + 1/4x , provethat cosec A + cot A = 2x or 1/2x

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Cosec(A) = x + 1/(4x)

now,

Cosec^2 (A) - Cot^2 (A) = 1,

so

Cot(A) = (+/-) Sqrt[ Cosec^2 (A) - 1]

so for our case:

Cot(A) = (+/-) Sqrt[ {x + 1/(4x)}^2 - 1 ]

= (+/-) Sqrt[ x^2 + 1/(4x)^2 +1/2 - 1 ]

= (+/-) Sqrt[ {x + 1/(4x)}^2 - 1/2 ]

= (+/-) Sqrt[ {x + 1/(4x)}^2 - 2.x.1/(4x) ]

= (+/-) Sqrt[ {x - 1/(4x)}^2 ]

=(+/-) {x - 1/(4x)} [for both x<1/4x & > 1/4x]

so let us use this for our expression:

Cosec(A) + Cot (A)

= {x + 1/(4x)} (+/-) {x - 1/(4x)}

= x + 1/(4x) + x - 1/(4x) [for + sign]

= 2 x

and

= x + 1/(4x) - x + 1/(4x) [for - sign]

= 2/(4 x)

=1/(2x)

so done.

The condition provided by the problem is vague since `csc A = (x+1)/(4x)` or `csc A = x + (1/4) x` .

Considering `csc A = (x+1)/(4x)` yields:

`csc A + cot A = 1/sin A + cos A/sin A = (1 + cos A)/sin A`

`1/sin A = (x+1)/(4x) =gt sin A = (4x)/(x+1)`

Using the basic formula of trigonometry yields:

`cos A = +-sqrt(1 - 16x^2/(x+1)^2) = gt cos A = +-sqrt((x+1-4x)(x+1+4x))/(x+1)`

`cos A = sqrt((1-3x)(1+5x))/(x+1) `

`csc A + cot A = (x + 1 + sqrt((1-3x)(1+5x)))/(4x)`

Notice that `csc A + cot A = (x + 1 + sqrt((1-3x)(1+5x)))/(4x) !=2x`

Considering `csc A =x + (1/4)x = 5x/4` yields:

`csc A + cot A = (1 + cos A)/sin A`

`csc A = 1/sin A => sin A = 4/(5x)`

`cos A = sqrt(1 - 16/(25x^2)) = sqrt(25x^2-16)/(5x)`

`csc A + cot A = (5x +sqrt(25x^2-16))/4 != 2x`

**Notice that either considering `csc A = x + (1/4)x ` or`csc A = (x+1)/(4x)` , the result is nor `2x` , neither `x/2` .**

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