2 tan x
1+ tan^{2 }x
2 tan x
1+ tan^{2 }x
Before solving this problem, it's useful to know how the three Pythagorean identities arise from the Pythagorean formula.
If you have a right triangle, select one of the non-right angles as x, and label the three sides as adj, opp, and hyp to represent the lengths of the sides adjacent to x, opposite from x, and the length of the hypotenuse. Then Pythagorean theorem states:
opp^{2} + adj^{2} = hyp^{2}
Divide both sides by hyp^{2} to obtain:
(opp/hyp)^{2} + (adj/hyp)^{2} = 1
By substituting sin x = opp/hyp and cos x = adj/hyp , we get the most important trigonometric identity:
sin^{2} x + cos^{2} x = 1.
If you've learned circular trigonometry then I recommend the website: http://www.mathsisfun.com/geometry/unit-circle.html for an interactive demonstration, and the site: http://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html has a great interactive unit circle.
I don't even bother memorizing the other two Pythagorean identities. I just derive them when I need them by dividing both sides of that most important trigonometric identity by cos^{2}
x or sin^{2} x. Dividing by cos^{2} x gives us:
(sin x / cos x)^{2} + 1 = (1 / cos x)^{2}
Using the definitions for tangent and secant in terms of sine and cosine yields:
tan^{2} x + 1 = sec^{2 }x
Of course, if you like memorization better than derivation then memorize all three identities, but you still should understand how to derive them. Using this identity with your problem, we get:
(2tan x) / (1+tan^{2} x) = 2tan x / sec^{2} x
= 2(sin x / cos x) / (1/cos^{2} x)
= 2(sin x / cos x)(cos^{2} x)
= 2 sin x cos x
That's a fairly simple expression, isn't it? But using other trigonometric identities that you're expected to know, you should be able to think of a way to simplify it even further.
From Pythagorean identities:
1 + tan^2 x = sec^2 x
and
tan^2 x = sin^2 x/cos^2 x
So, now instead of
2 tan x / (1 + tan^2 x)
we have:
2(sin^2 x/cos^2 x) / sec^2 x
but 1/cos^2 x = sec^2 x
2(sin^2 x *sec^2 x) / sec^2 x
2(sin^2 x)
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