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" cosec 2x + cot 2x = cot x "...

`" cosec 2x + cot 2x = cot x "`

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To solve the equation \( \csc 2x + \cot 2x = \cot x \), we will start with the left-hand side (LHS) and simplify it to show that it equals the right-hand side (RHS). ### Step 1: Write the LHS The left-hand side of the equation is: \[ \csc 2x + \cot 2x \] ...
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Knowledge Check

  • If cosec x = 1 = cot x, then x =

    A
    `n pi `
    B
    `2n pi`
    C
    `2n pi + pi/2`
    D
    `2n pi - pi/2`

  • If csc x = 1 + cot x , then x =

    A
    `n pi + (pi//2)`
    B
    `npi - (pi//2)`
    C
    `2n pi + (pi//2)`
    D
    `2n pi - (pi//2)`

  • If (cosec x + cot x)/(cosec x – cot x) = 7. Then, find the value of (4sin^2 x + 1)(4sin^2 x - 1)

    A
    `12//11`
    B
    `11//3`
    C
    `13//5`
    D
    `12//7`

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