How do you prove $\frac{cos [θ] + cot [θ]}{csc [θ] + 1} = cos [θ]$ $(cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]$ ?

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How do you prove $\frac{cos [θ] + cot [θ]}{csc [θ] + 1} = cos [θ]$ $(cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]$ ?

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Answer 1 · 2016-05-02T22:30:19

Using the definitions of $cot (θ)$ $cot(theta)$ and $csc (θ)$ $csc(theta)$ , for $sin (θ) \neq 0$ $sin(theta)!=0$ and $sin (θ) \neq - 1$ $sin(theta)!=-1$ , we have

$\frac{cos (θ) + cot (θ)}{csc (θ) + 1} = \frac{cos (θ) + \frac{cos (θ)}{sin (θ)}}{\frac{1}{sin (θ)} + 1}$ $(cos(theta)+cot(theta))/(csc(theta)+1) = (cos(theta)+cos(theta)/sin(theta))/(1/sin(theta)+1)$

$= cos (θ) \frac{1 + \frac{1}{sin (θ)}}{1 + \frac{1}{sin (θ)}}$ $=cos(theta)(1+1/sin(theta))/(1+1/sin(theta))$

$= cos (θ) \cdot 1$ $=cos(theta)*1$

$= cos (θ)$ $=cos(theta)$

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How do you prove $\frac{cos [θ] + cot [θ]}{csc [θ] + 1} = cos [θ]$ $(cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]$ ?

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How do you prove cos[θ]+cot[θ]csc[θ]+1=cos[θ](cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]?

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How do you prove $\frac{cos [θ] + cot [θ]}{csc [θ] + 1} = cos [θ]$ $(cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]$ ?