How to verify the identity: (1-2cos^2(x))/(sin(x)cos(x))?

How does one verify (1-2cos^2x)/(sinxcosx) = tanx-cotx using basic trig identities such as the Quotient Identities, the Reciprocal Identites, and the Pythagorean Identities?

3 Answers
Feb 25, 2018

tanx-cotx

Explanation:

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

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

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

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

sinx/(cosx)-cosx/(sinx)=

tanx-cotx

Feb 25, 2018

Please look at the Explanation area for this is a "how" question.

Explanation:

The first step to this problem is to use a Pythagorean Identity:

cos^2x=1-sin^2x

But we only want to replace one of the cos^2x so we can rewrite the identity like this for clarity:

(1−(cos^2x+cos^2x))/(sinxcosx)=tanx−cotx

And then complete the substitution:

(1−(1-sin^2x+cos^2x))/(sinxcosx)=tanx−cotx

Next, distribute the negative and simplify:

(sin^2x-cos^2x)/(sinxcosx)=tanx−cotx

Now we can split this fraction into two:

(sin^2x)/(sinxcosx)-(cos^2x)/(sinxcosx)=tanx−cotx

Simplify:

(sinx)/(cosx)-(cosx)/(sinx)=tanx−cotx

Apply the quotient identities:

(sinx)/(cosx) = tanx " and " (cosx)/(sinx) = cotx

And you reach:

tanx−cotx=tanx−cotx

Feb 25, 2018

"see explanation"

Explanation:

"using the "color(blue)"trigonometric identities"

•color(white)(x)tanx=sinx/cosx" and "cotx=cosx/sinx

.•color(white)(x)sin^2x+cos^2x=1

"consider the right side"

tanx-cotx

=sinx/cosx-cosx/sinx

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

=(1-cos^2x-cos^2x)/(sinxcosx)

=(1-2cos^2x)/(sinxcosx)=" left side "rArr" verified"