I am currently working on understanding trig identities. A question has me stumped, and no matter how I look at it, it never leads to the proof. I believe I am making a mistake when dividing multiple fractions.
$$ \frac{\csc x + \cot x}{\tan x + \sin x} = \cot x\csc x $$
For my first step I break up the $\csc x$ and $\cot x$ in the numerator and add them together to make:
$$\frac{\frac{1+\cos x}{\sin x\cos x}}{\tan x+\sin x}$$
I then simplify further and end up at:
$$ \frac{\cos x+\cos^2 x}{\sin^2 x\cos^2 x} $$
From here on I don't see any identities, or possible ways to decompose this further.
$\endgroup$ 12 Answers
$\begingroup$$\require{cancel}$As Chaitanya Tappu noted, you made a mistake when adding $\csc x$ and $\cot x$.
$$\frac{\csc x+\cot x}{\tan x+\sin x}=\frac{\frac{1}{\sin x}+\frac{\cos }{\sin x}}{\frac{\sin x}{\cos x}+\frac{\sin x\cos x}{\cos x}}=\frac{\frac{1+\cos x}{\sin x}}{\frac{\sin x(1+\cos x)}{\cos x}}=\frac{\cancel{1+\cos x}}{\sin x}\cdot\frac{\cos x}{\sin x\cancel{(1+\cos x)}}$$$$=\frac{\cos x}{\sin x}\cdot\frac{1}{\sin x}=\cot x\csc x$$
$\endgroup$ 1 $\begingroup$$$\dfrac{a+b}{\dfrac1a+\dfrac1b}=\cdots=ab$$ for $a+b\ne0$
$\tan x=\dfrac1?,\sin x=\dfrac1?$
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