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Verify the identity 1/tan(theta)csc(theta) = cos(theta)

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3 Answers

Use the identities tan(x) = sin(x)/cos(x) and csc(x) = 1/sin(x) to get:
 
1/(tan(x) × csc(x)) = 1/[(sin(x)/cos(x) × 1/sin(x)]
 
Now just work with the denominator (bottom) of the rational expression (fraction):
 
sin(x)/cos(x) × 1/sin(x) = 1/cos(x)
 
Now we have 
 
1/(1/cos(x) = 1 ÷ 1/cos(x) = 1× cos(x) = cos(x):
 1/ (tan A . CSC A)                    = Cos A
 
  ( Cos A / sinA) . 1/ CSCA          = COSA
 
    (COS A / SinA) .( Sin A/ 1)      =
 
          COSA                             = COSA
 
        
Verify the identity 1/tan(theta)csc(theta) = cos(theta).

My old trig teacher said to draw a vertical bar through the =; then nothing goes over the bar. You are NOT working with an equation. The idea is to use identities to transform a side into something ever closer to the other side. Usually you just work on one side; but you can work on both if you’re careful not to cross over the bar. (I's hard to draw a vertical line in this editor, so I'll use "=?" on each line instead.)

1/( tan(x) csc(x) ) =? cos(x)

On the left, substitute sin(x)/cos(x) for tan(x) and 1/sin(x) for csc(x):

1/( (sin(x)/cos(x))*(1/sin(x)) ) =? cos(x)

1/( (1/cos(x))*(sin(x)/sin(x)) ) =? cos(x)

1/( (1/cos(x))*(1) ) =? cos(x)

Just like 1/(1/2) = 2; the left side becomes:

cos(x), and it does = the right side, cos(x).
So we drop the "?" and assert:
cos(x) = cos(x) √