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# Prove that each identity holds for all permissible values of x.$\begin{array}{l}{\text { a) } 1+\cot ^{2} x=\csc ^{2} x} \\ {\text { b) } \tan x=\csc 2 x-\cot 2 x} \\ {\text { c) } \sec x+\tan x=\frac{\cos x}{1-\sin x}} \\ {\text { d) } \frac{1}{1+\cos x}+\frac{1}{1-\cos x}=2 \csc ^{2} x}\end{array}$

Solution

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#### a)

Starting from the left hand side we get

$1+\cot^2 x=1+\frac{\cos^2 x}{\sin^2 x}=\frac{1}{\sin^2 x}=\csc^2 x$

which is what we wanted to prove.

#### b)

Starting from the right hand side we get

\begin{align*} \csc 2x-\cot 2x&=\frac{1}{\sin 2x}-\frac{\cos 2x}{\sin 2x}\\ &=\frac{1-\cos 2x}{\sin 2x}\\ &=\frac{1-(\cos^2 x-\sin^2 x)}{2\sin x\cos x}\\ &=\frac{\sin^2 x+\sin^2 x}{2\sin x\cos x}\\ &=\frac{2\sin^2 x}{2\sin x\cos x}\\ &=\frac{\sin x}{\cos x}\\ &=\tan x \end{align*}

which is what we wanted to show.

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