Question

PROVE √(1-〖cos〗^2 θ)/cosθ=tanθ

Answer 1

Explanation:

`(√(1-\cos^2\theta))/(\cos\theta)=\tan\theta\\\\\text{We know:}\\(1)\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\(2)\tan x=(\sin x)/(\cos x)\\\\\text{For}\ √(1-\cos^2\theta)\ \text{use}\ (1):\\\\L_S=(√(1-\cos^2\theta))/(\cos\theta)=(√(\sin^2\theta))/(\cos\theta)=(\sin\theta)/(\cos\theta)=\tan\theta\ \ \ _(used (2))\\\\R_S=\tan\theta\\\\L_S=R_S`

`\text{Of course, this equality is not true for everyone values of}\ \theta.\\\\1^o.\ \cos\theta\\eq0\to\theta\\eq(\pi)/(2)+k\pi,\ k\in\mathbb{Z}\\\\2^o.\ √(\sin^2\theta)=|\sin\theta|,\ \text{therefore}\\\\\text{for}\ \theta\in[2k\pi;\ (2k+1)\pi],\ \sin\theta\geq0\to|\sin\theta|=\sin\theta\\\\\text{therefore}\ (|\sin\theta|)/(\cos\theta)=(\sin\theta)/(\cos\theta)=\tan\theta`

`\text{for}\ \theta\in\bigg((2k-1)\pi;\ 2k\pi\bigg),\ \sin\theta<0\to|\sin\theta|=-\sin\theta\\\\\text{therefore}\ (|\sin\theta|)/(\cos\theta)=(-\sin\theta)/(\cos\theta)=-\tan\theta\\eq\tan\theta`

`\text{Conclusion:}\\\\\text{This equality is true for}\ \theta\in\left[2k\pi;\ (\pi)/(2)+2k\pi\right)\cup\left((\pi)/(2)+2k\pi;\ \pi+2k\pi\right],\ k\in\mathbb{Z}`

Answer 2

The answer is below.

Explanation:

PROVE √(1-〖cos〗^2 θ)/cosθ = tanθ

Now 1 - cos^2 θ = sin^2θ

So √(1 - cos^2 θ) = sin θ

and √(1 - cos^2 θ) / cos θ

= sin θ / cos θ

= tan θ