If 1−sin⁡x1+sin⁡x=4, what is the value of tan⁡x2? 1)-3 2)2 3)-2 4)3

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If 1−sin⁡x1+sin⁡x=4, what is the value of tan⁡x2?  1)-3   2)2   3)-2   4)3

Bertha Jordan · Accepted Answer

Let t be t=tan⁡(x2) for the &quot;good x&quot; satisfying the given relation. Then sin⁡x=2t1+t2, so  4=1−sin⁡x1+sin⁡x=(1+t2)−2t(1+t2)+2t=(1−t1+t)2  This gives for 1−t1+t the values ±2, leading to the two solutions -3 and -1/3

maul124uk · Accepted Answer

You have  1−sin⁡x1+sin⁡x=4⇔sin2(x2)+cos2(x2)−2sin⁡(x2)cos⁡(x2)sin2(x2)+cos2(x2)+2sin⁡(x2)cos⁡(x2)=4  ⇔(sin⁡(x2)−cos⁡(x2)sin⁡(x2)+cos⁡(x2))2=4  ⇔sin⁡(x2)−cos⁡(x2)sin⁡(x2)+cos⁡(x2)=±2  ⇔tan⁡(x2)−1tan⁡(x2)+1=±2  The only solution of the equation tan⁡(x2)−1tan⁡(x2)+1=2 is tan⁡(x2)=−3 which is on that list, whereas the only solution of the equation tan⁡(x2)−1tan⁡(x2)+1=−2 is tan⁡(x2)=−13, which is not on that list.  So, the problem has two solutions, but only one of them is on the list of options.

karton · Accepted Answer

tan⁡x2=sin⁡x2cos⁡x2=2sin⁡x2cos⁡x22cos2⁡x2=sin⁡xcos⁡x+1 hence with sin⁡x=−35, cos⁡x=±1−sin2⁡x=±45, tan⁡x2=−13, −3

If \frac{1-\sin x}{1+\sin x}=4, what is the value of \tan

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