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

How does one verify (1-2cos^2x)/(sinxcosx) = tanx-cotx12cos2xsinxcosx=tanxcotx using basic trig identities such as the Quotient Identities, the Reciprocal Identites, and the Pythagorean Identities?

3 Answers
Feb 25, 2018

tanx-cotxtanxcotx

Explanation:

(1-2cos^2(x))/(sin(x)cos(x))=12cos2(x)sin(x)cos(x)=

((sin^2(x)+cos^2(x))-2cos^2(x))/(sin(x)cos(x))=(sin2(x)+cos2(x))2cos2(x)sin(x)cos(x)=

(sin^2(x)-cos^2(x))/(sin(x)cos(x))=sin2(x)cos2(x)sin(x)cos(x)=

sin^2x/(sinxcosx)-cos^2x/(sinxcosx)=sin2xsinxcosxcos2xsinxcosx=

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

tanx-cotxtanxcotx

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^2xcos2x=1sin2x

But we only want to replace one of the cos^2xcos2x so we can rewrite the identity like this for clarity:

(1−(cos^2x+cos^2x))/(sinxcosx)=tanx−cotx1(cos2x+cos2x)sinxcosx=tanxcotx

And then complete the substitution:

(1−(1-sin^2x+cos^2x))/(sinxcosx)=tanx−cotx1(1sin2x+cos2x)sinxcosx=tanxcotx

Next, distribute the negative and simplify:

(sin^2x-cos^2x)/(sinxcosx)=tanx−cotxsin2xcos2xsinxcosx=tanxcotx

Now we can split this fraction into two:

(sin^2x)/(sinxcosx)-(cos^2x)/(sinxcosx)=tanx−cotxsin2xsinxcosxcos2xsinxcosx=tanxcotx

Simplify:

(sinx)/(cosx)-(cosx)/(sinx)=tanx−cotxsinxcosxcosxsinx=tanxcotx

Apply the quotient identities:

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

And you reach:

tanx−cotx=tanx−cotxtanxcotx=tanxcotx

Feb 25, 2018

"see explanation"see explanation

Explanation:

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

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

.•color(white)(x)sin^2x+cos^2x=1xsin2x+cos2x=1

"consider the right side"consider the right side

tanx-cotxtanxcotx

=sinx/cosx-cosx/sinx=sinxcosxcosxsinx

=(sin^2x-cos^2x)/(sinxcosx)=sin2xcos2xsinxcosx

=(1-cos^2x-cos^2x)/(sinxcosx)=1cos2xcos2xsinxcosx

=(1-2cos^2x)/(sinxcosx)=" left side "rArr" verified"=12cos2xsinxcosx= left side verified