Show that $(2cos^2(x/2)tanx)/(tanx) = (tanx+cosxtanx)/(tanx)$ ?

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Answer 1 · 2015-10-24T00:49:14

Okay, let's consider dividing both sides by $tanx$ to begin with:

$(2cos^2(x/2)cancel(tanx))/cancel(tanx) = (cancel(tanx)(1 + cosx))/cancel(tanx)$

$2cos^2(x/2) = 1 + cosx => 2cos^2(x/2) - 1 = cosx$

Notice how we can use the identity $sin^2(u) + cos^2(u) = 1$ , where $u = x/2$ :

$2cos^2(x/2) - [sin^2(x/2) + cos^2(x/2)] = cosx$

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

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

Then, we can use the identity $cos(u + v) = cosucosv - sinusinv$ , where $u = v = x/2$ and work backwards:

$cos(x/2)cos(x/2) - sin(x/2)sin(x/2) = cosx$

$cos(x/2 + x/2) = cosx$

$color(blue)(cosx = cosx)$

Show that (2cos^2(x/2)tanx)/(tanx) = (tanx+cosxtanx)/(tanx)(2cos^2(x/2)tanx)/(tanx) = (tanx+cosxtanx)/(tanx)?