Trigonometric identity?

Question

I need help on establishing a trigonometric identity for this problem:

csc (theta) - 1 / cot (theta) = cot (theta) / csc (theta) + 1

Thanks!

Answer 1

(csc(t) - 1) / cot(t) = cot(t) / (csc(t) + 1)

Choose the right hand side, RHS.

RHS = cot(t) / (csc(t) + 1)

Change everything to sines and cosines.

RHS = [cos(t)/sin(t)] / [ 1/sin(t) + 1]

Get rid of the complex fraction by multiplying numerator and denominator by sin(t).

RHS = [cos(t)] / [1 + sin(t)]

Multiply the numerator and denominator by the conjugate of the denominator. Note that the conjugate of (a + b) is (a - b), and when multiplied together, we get a difference of squares.

RHS = [cos(t)][1 - sin(t)] / [1 - sin^2(t)]
RHS = [cos(t)][1 -sin(t)] / cos^2(t)

And we get our cosine on the numerator to cancel.

RHS = [1 - sin(t)]/[cos(t)]

Let's go with the LHS now.

LHS = (csc(t) - 1) / cot(t)

Change to sines and cosines.

LHS = (1/sin(t) - 1) / (cos(t)/sin(t))

Multiply top and bottom by sin(t)

LHS = (1 - sin(t)) / cos(t)

LHS = RHS

Answer 2

(csc(theta)-1) / cot(theta)

Multiply top and bottom by (csc(theta)+1)

(csc(theta)-1)(csc(theta)+1) / (cot(theta) (csc(theta)+1))

(csc^2(theta)-1) / (cot(theta) (csc(theta)+1))

cot^2(theta) / (cot(theta) (csc(theta)+1))

cot(theta) / (csc(theta)+1)