Question

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$\sf cos ^2+ tan^2=1$

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Answer 1

Use double angle formula to remove coefficient inside the cos, then rearrange standard trig definitions to make the trig function match the inverse trig function inside the bracket

Explanation:

Recall the double angle formula:

$cos2\theta=1-2\sin^2\theta$

Then  cos$(2\arctanx)=1-2\sin^2\arctanxcos$ NB I've written "arctan" here rather than "

tan $tan^(-1)$because the combination of exponents meaning powers and function inverses is potentially confusing.

So we now have a trig function of an inverse trig function. If we can express our

because the combination of exponents meaning powers and function inverses is potentially confusing.

So we now have a trig function of an inverse trig function. If we can express our  sin in terms of  tan this will cancel right out sin in terms of tan, this will cancel right out.

By definition, tan θ$\tan\theta=(\sin\theta)/(cos\theta)=(sin\theta)/\sqrt(1-sin^2\theta)$

so

$\tan^2\theta(1-\sin^2\theta)=\sin^2\theta$

$\tan^2\theta=\sin^2\theta(1+\tan^2\theta)$

$sin^2\theta=tan^2\theta/(1+tan^2\theta)$

By definition

$\\\\\\\\\\\\tan\arctan x =$x

so $1-2\sin^2\arctanx$ becomes 1−$(2x^2)/(1+x^2)$

Putting this over a common denominator makes $(1-x^2)/(1+x^2)$

So

$\cos(2\arctanx)=(1-x^2)/(1+x^2)$

Answer 2

Use double angle formula to remove coefficient inside the cos, then rearrange standard trig definitions to make the trig function match the inverse trig function inside the bracket

Explanation:

Recall the double angle formula:

−1) because the combination of exponents meaning powers and function inverses is potentially confusing.

So we now have a trig function of an inverse trig function. If we can express our

because the combination of exponents meaning powers and function inverses is potentially confusing.

So we now have a trig function of an inverse trig function. If we can express our sin in terms of tan this will cancel right out sin in terms of tan, this will cancel right out.

By definition, tan θ\tan\theta=(\sin\theta)/(cos\theta)=(sin\theta)/\sqrt(1-sin^2\theta)tanθ=(sinθ)/(cosθ)=(sinθ)/

(

By definition

$\begin{gathered}\\\\\\\\\\\\tan\arctan x =\end{gathered} </p><p>tanarctanx=$

x

so 1-2\sin^2\arctanx1−2sin arctanx becomes 1−(2x^2)/(1+x^2)(2x

Putting this over a common denominator makes (1-x^2)/(1+x^2)(1−x