Question

Can someone pls do 25 and 26 with working

Answer 1

25.

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$\dfrac{\tan x + \sin x}{\tan x - \sin x} = \dfrac{\sec x + 1}{\sec x - 1}$

$\dfrac{\tan x + \sin x}{\tan x - \sin x}$

$= \dfrac{\frac{\sin x}{\cos x} + \sin x}{\frac{\sin x}{\cos x} - \sin x}$

$= \dfrac{\sin x ( \frac{1}{\cos x} + 1)}{\sin x( \frac{ 1}{\cos x} - 1)}$

$= \dfrac{\sec x + 1}{\sec x - 1} \quad\checkmark$

26.

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$\dfrac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta +1} = \dfrac{1+\cos \theta}{\sin \theta}$

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 $\theta=2x$.

$\dfrac{\cot 2x + \csc 2x - 1}{\cot 2x- \csc 2x +1}$

$=\dfrac{ (\cos 2x / \sin 2x) + (1/\sin 2x ) - 1}{\cos 2x/\sin 2x - 1/\sin 2x + 1}$

$=\dfrac{\cos 2x + 1 - \sin 2x}{\cos2x - 1 + \sin 2x}$

$=\dfrac{2\cos^2 x - 1 + 1 -2 \cos x \sin x }{1 - 2\sin^2x-1 + 2\sin x \cos x}$

$=\dfrac{2 \cos^2 x -2 \cos x \sin x }{-2\sin^2x+2 \sin x \cos x}$

$=\dfrac{2\cos x}{2 \sin x} \cdot \dfrac{\cos x- \sin x}{-\sin x +\cos x}$

$=\dfrac{2\cos x}{2\sin x}$

$=\dfrac{2\cos^2 x}{2\sin x\cos x}$

$=\dfrac{1 + 2\cos^2 x - 1}{2\sin x\cos x}$

$= \dfrac{1 + \cos 2x}{\sin 2x}$

$=\dfrac{1+\cos \theta}{\sin \theta} \quad\checkmark$

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For another answer, let's use the hint on this one, which was to write

$1=\csc^2 \theta - \cot^2 \theta$

That's a good hint; first let's verify if it's true.

$\sin^2 \theta + \cos^2 \theta = 1$

$\sin^2 \theta = 1 - \cos^2 \theta$

$1 = \dfrac{1}{\sin ^2 \theta} - \dfrac{\cos ^2 \theta}{\sin ^2 \theta}$

$1 = \csc^2 \theta - \cot^2 \theta \quad\checkmark$

Now,

$\dfrac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta +1}$

$= \dfrac{\cot \theta + \csc \theta - (\csc^2 \theta - \cot^2 \theta)}{\cot \theta - \csc \theta +1}$

$= \dfrac{\cot \theta + \csc \theta +(\cot^2 \theta - \csc^2 \theta)}{\cot \theta - \csc \theta +1}$

$= \dfrac{\cot \theta + \csc \theta +(\cot \theta - \csc \theta)(\cot \theta + \csc \theta)}{\cot \theta - \csc \theta +1}$

$= \dfrac{(\cot \theta + \csc \theta)(1+ \cot \theta - \csc \theta)}{\cot \theta - \csc \theta +1}$

$=\cot \theta + \csc \theta$

$=\dfrac{ \cos \theta}{\sin \theta}+ \dfrac{1}{\sin \theta}$

$=\dfrac{1+ \cos \theta}{\sin \theta} \quad\checkmark$