Question #74e29

1 Answer
Jim H · EZ as pi
Mar 2, 2017

When in doubt, rewrite using sine and cosine. See below.

Explanation:

#(sin^2(theta)-tan(theta)) / (cos^2(theta)-cot(theta)) = (sin^2(theta)-sin(theta)/cos(theta)) / (cos^2(theta)-cos(theta)/sin(theta))#

Factor out #sintheta# on top and #cos theta# on the bottom, to get a #tantheta#

# = (sin(theta)(sin(theta)-1/(cos(theta))))/(cos(theta)(cos(theta)-1/sin(theta)))#

For the remaining quotient, get a single fraction over another fraction. (No I'm not certain it will work, but I have to try something, so . . . )

# = tan(theta) (((sin(theta)cos(theta)-1)/(cos(theta))))/(((sin(theta)cos(theta)-1)/(sin(theta))))#

We now see that we have the same numerators, so when we invert the denominator and multiply, the #((sin(theta)cos(theta)-1)# will cancel and leave #sintheta/costheta#

# = tan(theta)((sin(theta)cos(theta)-1)/(cos(theta)) * sin(theta)/((sin(theta)cos(theta)-1)))#

# = tan(theta) sin(theta)/cos(theta)#

# = tan^2(theta)#

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