(20secqtanq) / (10-secq)^2

Question

how do I simplify this problem using trig identities

 

Answer 1

 

 20 SECq * Tan q / ( 10 - sec q ) ²

 

  ( 20 / Cosq * Sinq/ Cos q )/ ( 10 - 1/ cos q) ^{2 =}

 

  ( 20 sin q / cos ² q ) / ( 10 Cosq - 1 / Cos q ) ² = 

 

 ( 20 Sin q Cos² q ) / Cos² q ( 10Cosq - 1 )²  = 

    

    ( 20 Sin q ) / ( 10 Cos q - 1 ) ^{2 =}

 

    

Answer 2

You can get the answer by multiplying the top and the bottom by (cosq)^2.

 

 

 

(20 Sec q Tan q)(cosq)^2

 

  (10 - Sec q)²(cosq)^2

 

 

=      20 sinq    

 

  (10cosq - 1)²

 

Answer 3

William and Daisy,

 

I just worked it out and got the same answer.  Give me a minute and I will show the work.

 

Bill

 

 

Answer 4

[20*sec(q)*tan(q)]

_______________

[10 - sec(q)]²

 

sec(q) = 1/cos(q); tan(q) = sin(q)/cos(q)

 

sec(q)*tan(q) = 1/cos(q) * [sin(q)/cos(q)] = sin(q)/cos²(q) = sin(q)/[1-sin²(q)]

 

      20*sin(q)

=  ______________

    [10*cos(q) -1]²

 

Daisy, the answer in bold type is what my TI-89 Titanium calculator came back with when I input the original expression, but I have no idea how to "prove" it.