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$\large{\text{cosec }\theta+\sec{\theta}=\dfrac{\sin\theta+\cos\theta}{\sin\theta\,\cos\theta}}$

I know that cosecant is the inverse of sine, and secant is the inverse of cosine. However, that does not equal the right hand side of the equation. I know nothing about what to do next.

Semiclassical
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xsqs
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    Actually, the cosecant is not the inverse of the sine, but rather its reciprocal. Similarly, the secant is not the inverse of the cosine, but rather its reciprocal. – Senex Ægypti Parvi Aug 07 '14 at 01:34
  • Can you show that $\frac1a +\frac1b = \frac{a +b}{ab}$? Because it's exactly the same thing. – MPW Aug 07 '14 at 02:38

3 Answers3

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Perhaps most algebraically natural is to go from right to left. We have $$\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac{\sin\theta}{\sin\theta\cos\theta}+\frac{\cos\theta}{\sin\theta\cos\theta}=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}=\sec\theta+\csc\theta.$$

However, going from left to right is also in a certain sense natural. Express the left side in terms of sines and cosines. We have $$\csc\theta+\sec\theta=\frac{1}{\sin\theta}+\frac{1}{\cos\theta}.$$ Now bring the expression on the right to a common denominator $\sin\theta\cos\theta$.

André Nicolas
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$\begin{align}\csc\theta + \sec\theta & = \frac{1}{\sin\theta}+\frac{1}{\cos\theta} \\[2.5ex] & = \frac{\cos\theta}{\sin\theta\cos\theta}+\frac{\sin\theta}{\sin\theta\cos\theta} \\[2.5ex] & =\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\end{align}$

JB-Franco
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please delete me
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Substitute $1\over{\cos{\theta}}$ for $\sec{\theta}$ and $1\over{\sin{\theta}}$ for $\csc{\theta}$. You can take it from there.

Senex Ægypti Parvi
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