can someone pls do 25 and 26 with working

25.Show[tex] \dfrac{\tan x + \sin x}{\tan x - \sin x} = \dfrac{\sec x + 1}{\sec x - 1}[/tex][tex] \dfrac{\tan x + \sin x}{\tan x - \sin x}[/tex][tex]= \dfrac{\frac{\sin x}{\cos x} + \sin x}{\frac{\sin x}{\cos x} - \sin x}[/tex][tex]= \dfrac{\sin x ( \frac{1}{\cos x} + 1)}{\sin x( \frac{ 1}{\cos x} - 1)}[/tex][tex]= \dfrac{\sec x + 1}{\sec x - 1} \quad\checkmark[/tex]26.Show [tex]\dfrac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta +1} = \dfrac{1+\cos \theta}{\sin \theta}[/tex]That's the cotangent half angle formula on the right, so I guess on the left too.I fooled around with this one for a while before I took the hint which was to let [tex]\theta=2x[/tex].[tex]\dfrac{\cot 2x + \csc 2x - 1}{\cot 2x- \csc 2x +1}[/tex][tex]=\dfrac{ (\cos 2x / \sin 2x) + (1/\sin 2x ) - 1}{\cos 2x/\sin 2x - 1/\sin 2x + 1}[/tex][tex]=\dfrac{\cos 2x + 1 - \sin 2x}{\cos2x - 1 + \sin 2x}[/tex][tex]=\dfrac{2\cos^2 x - 1 + 1 -2 \cos x \sin x }{1 - 2\sin^2x-1 + 2\sin x \cos x}[/tex][tex]=\dfrac{2 \cos^2 x -2 \cos x \sin x }{-2\sin^2x+2 \sin x \cos x}[/tex][tex]=\dfrac{2\cos x}{2 \sin x} \cdot \dfrac{\cos x- \sin x}{-\sin x +\cos x}[/tex][tex] =\dfrac{2\cos x}{2\sin x}[/tex][tex] =\dfrac{2\cos^2 x}{2\sin x\cos x}[/tex][tex] =\dfrac{1 + 2\cos^2 x - 1}{2\sin x\cos x}[/tex][tex] = \dfrac{1 + \cos 2x}{\sin 2x}[/tex][tex]=\dfrac{1+\cos \theta}{\sin \theta} \quad\checkmark [/tex]------For another answer, let's use the hint on this one, which was to write [tex]1=\csc^2 \theta - \cot^2 \theta[/tex]That's a good hint; first let's verify if it's true.[tex]\sin^2 \theta + \cos^2 \theta = 1[/tex][tex]\sin^2 \theta = 1 - \cos^2 \theta[/tex][tex]1 = \dfrac{1}{\sin ^2 \theta} - \dfrac{\cos ^2 \theta}{\sin ^2 \theta}[/tex][tex]1 = \csc^2 \theta - \cot^2 \theta \quad\checkmark[/tex]Now,[tex]\dfrac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta +1}[/tex][tex]= \dfrac{\cot \theta + \csc \theta - (\csc^2 \theta - \cot^2 \theta)}{\cot \theta - \csc \theta +1}[/tex][tex]= \dfrac{\cot \theta + \csc \theta +(\cot^2 \theta - \csc^2 \theta)}{\cot \theta - \csc \theta +1}[/tex][tex]= \dfrac{\cot \theta + \csc \theta +(\cot \theta - \csc \theta)(\cot \theta + \csc \theta)}{\cot \theta - \csc \theta +1}[/tex][tex]= \dfrac{(\cot \theta + \csc \theta)(1+ \cot \theta - \csc \theta)}{\cot \theta - \csc \theta +1}[/tex][tex]=\cot \theta + \csc \theta[/tex][tex]=\dfrac{ \cos \theta}{\sin \theta}+ \dfrac{1}{\sin \theta}[/tex][tex]=\dfrac{1+ \cos \theta}{\sin \theta} \quad\checkmark[/tex]