Prove the following identity :Sin α . Cos α .
Tan α = (1 – Cos
α) (1 + Cos
α)
2 answers:
<span><u><em>Question </em></u> Sin α . Cos α .
Tan α = (1 – Cos
α) (1 + Cos
α) <u><em>Answer </em></u> <u><em>Left side </em> </u> = Sin β . Tan β
+ Cos β = Sin </span>β . Sin β / Cos β + Cos β = Sin² β / Cos β + Cos² β / Cos β = 1 / Cos β = Sec β = <u><em>Right side proven </em></u>
Let's work on the left side first. And remember that the<u> tangent</u> is the same as <u>sin/cos</u>. sin(a) cos(a) tan(a) Substitute for the tangent: [ sin(a) cos(a) ] [ sin(a)/cos(a) ] Cancel the cos(a) from the top and bottom, and you're left with [ sin(a) ] . . . . . [ sin(a) ] which is [ <u>sin²(a) </u> ] That's the <u>left side</u>. Now, work on the right side: [ 1 - cos(a) ] [ 1 + cos(a) ] Multiply that all out, using FOIL: [ 1 + cos(a) - cos(a) - cos²(a) ] = [ <u>1 - cos²(a)</u> ] That's the <u>right side</u>. Do you remember that for any angle, sin²(b) + cos²(b) = 1 ? Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle. So, on the <u>right side</u> , you could write [ <u>sin²(a)</u> ] . Now look back about 9 lines, and compare that to the result we got for the <u>left side</u> . They look quite similar. In fact, they're identical. And so the identity is proven. Whew !
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