Trigonometric identities?

1) cot(x) - tan(x)

= (cos(x)/sin(x)) - (sin(x)/cos(x))

= (cos^2(x) - sin^2(x)) / sin(x)cos(x)

= cos(2x) / sin(x)cos(x)

= cos(2x) / (1/2)sin(2x)

= (1 / (1/2))(cos(2x)/sin(2x))

= 2 cot(2x).

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2) If the first term is supposed to be 1/(cosec^2(x) - 1), this is what you get:

(1 / (cosec^2(x) - 1)) + sin^2(x) + cos^2(x)

= (1/cot^2(x)) + sin^2(x) + cos^2(x)

= (1/cot^2(x)) + 1

= tan^2(x) + 1

= sec^2(x).

If instead the first term is as you typed, follow Wayne's steps.

= cot(x) - tan(x) → you know that: cot(x) = cos(x)/sin(x)

= [cos(x)/sin(x)] - tan(x) → you know that: tan(x) = sin(x)/cos(x)

= [cos(x)/sin(x)] - [sin(x)/cos(x)]

= [cos²(x) + sin²(x)]/[sin(x).cos(x)] → recall: cos²(x) + sin²(x) = 1

= 1/[sin(x).cos(x)] → to go further

= [1/sin(x)].[1/cos(x)]

= csc(x).sec(x)

= {1/[csc²(x) - 1]} + sin²(x) + cos²(x) → recall: cos²(x) + sin²(x) = 1

= {1/[csc²(x) - 1]} + 1

= {1 + [csc²(x) - 1]}/[csc²(x) - 1]

= {1 + csc²(x) - 1}/[csc²(x) - 1]

= csc²(x)/[csc²(x) - 1] → you know that: csc(x) = 1/sin(x)

= {1/sin²(x)]/[{1/sin²(x)} - 1]

= {1/sin²(x)]/[{1 - sin²(x)}/sin²(x)]

= 1/[1 - sin²(x)] → you know that: cos²(x) + sin²(x) = 1

= 1/[cos²(x) + sin²(x) - sin²(x)]

= 1/cos²(x)

= sec²(x)

1) cotx => cosx/sinx

tanx => sinx/cosx

so, cotx - tanx = cosx/sinx - sinx/cosx

i.e. (cos²x - sin²x)/sinxcosx

=> cos2x/sinxcosx

or, 2cos2x/2sinxcosx

i.e. 2cos2x/sin2x

=> 2cot2x

2) I will assume that by (1/cosec²x - 1) you mean:

(1/cosec²x) - 1

i.e. (1/(1/sin²x)) - 1

=> sin²x - 1

so, sin²x - 1 + sin²x + cos²x

or, sin²x - 1 + 1

Hence, sin²x

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