Question

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?

Answer 1

`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`

Answer 2

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`

Answer 3

`"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"`