I have been stuck with this part of question and need help:

Prove that: $$ \frac{1}{\left(\tan\frac{x}{2}+1\right)^2\cos^2\frac{x}{2}} = \frac{1}{1+\sin x}$$

Thank you!

  • 1
    $\begingroup$ Of course you can immediately take reciprocals of both sides. Do you know the half-angle formulas? $\endgroup$ – Greg Martin Mar 26 at 1:16
  • $\begingroup$ Yes, I tried using them but looks like it went the wrong way. Thanks anyway! $\endgroup$ – FAF Mar 26 at 1:39

$$\frac{1}{(\tan\frac{x}{2}+1)^2\cos^2\frac{x}{2}} =\frac{1}{(\tan^2\frac{x}{2}+2\tan\frac{x}{2}+1)\cos^2\frac{x}{2}}$$ $$=\frac{1}{\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}+\cos^2\frac{x}{2}}=\frac{1}{1+2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{1}{1+\sin x}$$


Let $\tan\dfrac x2=t$

So we have $$\dfrac{1+t^2}{(t+1)^2}=\cdots=\dfrac1{1+\dfrac{2t}{1+t^2}}$$

Now use

  • $\begingroup$ (+) nice........ $\endgroup$ – E.H.E Mar 26 at 2:20

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