Answer:
See Below.
Explanation:
We want to verify the identity:
![\displaystyle \csc^2 x -2\csc x \cot x +\cot ^2 x = \tan^2\left((x)/(2)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/y4p0ney9jitu3760xbts07svq9sjzl7kgi.png)
Note that the left-hand side is a perfect square trinomial pattern. Namely:
![a^2-2ab+b^2=(a-b)^2](https://img.qammunity.org/2022/formulas/mathematics/high-school/btpswuymiedkig9ld717rq30wmkabishnd.png)
If we let a = csc(x) and b = cot(x), we can factor it as such:
![\displaystyle (\csc x - \cot x)^2 = \tan^2\left((x)/(2)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/jpf93vxd2yecwid8fjm2ejglo64c2e4kyn.png)
Let csc(x) = 1 / sin(x) and cot(x) = cos(x) / sin(x):
![\displaystyle \left((1)/(\sin x)-(\cos x )/(\sin x)\right)^2=\tan^2\left((x)/(2)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/iegz9k9w60tkp319stxmvkvzv2gbgojniq.png)
Combine fractions:
![\displaystyle \left((1-\cos x)/(\sin x)\right)^2=\tan^2\left((x)/(2)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/bwmqhysd5yrj1r1beckl6juff38jt4xxju.png)
Square (but do not simplify yet):
![\displaystyle ((1-\cos x)^2)/(\sin ^2x)=\tan^2\left((x)/(2)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/buom6lylf1o1650q3xwkz4i67j7wn1m6iv.png)
Now, we can make a substitution. Let u = x / 2. So, x = 2u. Substitute:
![\displaystyle ((1-\cos 2u)^2)/(\sin ^22u)=\tan^2u](https://img.qammunity.org/2022/formulas/mathematics/high-school/6csqo45f4ch8l1ttvp1jbsunxeqcoegvoh.png)
Recall that cos(2u) = 1 - sin²(u). Hence:
![\displaystyle ((1-(1-2\sin^2u))^2)/(\sin ^2 2u)=\tan^2u](https://img.qammunity.org/2022/formulas/mathematics/high-school/3ftrrb0mgt5lbm7ld1ov4g0ekkhcfz9e55.png)
Simplify:
![\displaystyle (4\sin^4 u)/(\sin ^2 2u)=\tan^2 u](https://img.qammunity.org/2022/formulas/mathematics/high-school/xlsq23tl643rescnk2f486rqi1ngdet4vb.png)
Recall that sin(2u) = 2sin(u)cos(u). Hence:
![\displaystyle (4\sin^4 u)/((2\sin u\cos u)^2)=\tan^2 u](https://img.qammunity.org/2022/formulas/mathematics/high-school/7e8gleocmsgtye2lt0sc86rjfjyqsghdfk.png)
Square:
![\displaystyle (4\sin^4 u)/(4\sin^2 u\cos ^2u)=\tan^2 u](https://img.qammunity.org/2022/formulas/mathematics/high-school/fx2idfi9o7x1ckpau63izc4o0ng6uui305.png)
Cancel:
![\displaystyle (\sin ^2 u)/(\cos ^2 u)=\tan ^2 u](https://img.qammunity.org/2022/formulas/mathematics/high-school/oihiatqbv4uo4247rsh1xax91w6lwb66db.png)
Since sin(u) / cos(u) = tan(u):
![\displaystyle \left((\sin u)/(\cos u)\right)^2=\tan^2u=\tan^2u](https://img.qammunity.org/2022/formulas/mathematics/high-school/w3i9sk232djx8izb0a0wzndcp6obej659r.png)
We can substitute u back for x / 2:
![\displaystyle \tan^2\left((x)/(2)\right)= \tan^2\left((x)/(2)\right)](https://img.qammunity.org/2022/formulas/mathematics/high-school/tdabk9lq22oloqy9g6xy0uzpcl00djmes3.png)
Hence proven.