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Can anyone help me solve the following trig equations.

$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$$

My work thus far

$$\frac{\frac{1}{\cos{A}}+\frac{1}{\sin{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$

$$\frac{\frac{\sin{A} + \cos{A}}{\sin{A} * \cos{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$

But how would I continue?

My second question is

$$\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$

My work is

$$\frac{\cos{A}}{\sin{A}} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$

I think I know how to solve this one by using a common denominator but I am not sure.

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Once you obtain ${\sin A+\cos A\over\sin A\cos A}\over{\sin A\over\cos A}+{\cos A\over\sin A}$, multiply top and bottom by $\sin A\cos A$ (or, otherwise, simplify the fraction by first doing the addition downstairs). –  David Mitra Jul 16 '12 at 21:34

3 Answers 3

Try waiting until the last minute before converting to sines and cosines.

$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}$$ Recall that $\tan A\cot A = 1$ and $\tan A = \frac{\sec A}{\csc A}$

$$=\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}\cdot\frac{\tan A}{\tan A} $$ $$=\frac{\sec{A}\tan A+\csc{A}\tan A}{\tan^2{A} + 1} $$ $$=\frac{\sec{A}\tan A+\csc{A}\cdot\frac{\sec A}{\csc A}}{\sec^2 A} $$ $$=\frac{\sec{A}\tan A+\sec A}{\sec^2 A} $$ $$=\frac{\sec A (\tan A+1)}{\sec^2 A} $$ $$=\frac{\tan A+1}{\sec A} $$ $$=\frac{\frac{\sec A}{\csc A}}{\sec A} + \frac{1}{\sec A} $$ $$=\frac{1}{\csc A} + \frac{1}{\sec A} $$ $$=\sin A + \cos A $$

But I do like hjpotter92's answer better :)

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Solution 1:

$$\dfrac{\dfrac{\sin{A} + \cos{A}}{\sin{A} \cos{A}}}{\dfrac{\sin^2{A} + \cos^2{A}}{\sin{A} \cos{A}}}$$

$$ = \frac{\sin{A} + \cos{A}}{\sin^2{A} + \cos^2{A}}$$

$$ = \sin{A} + \cos{A}$$

Solution 2:

$$\frac{\cos{A}(1 + \cos{A}) + \sin^2{A}}{\sin{A} (1 + \cos{A})}$$

$$= \frac{\color{red}{\cos{A} + 1}}{\sin{A} (\color{red}{\cos{A} + 1})}$$

$$= \frac{1}{\sin{A}} = \csc{A}$$

PS: I don't know how to put those cross-marks(cancellations) on fractions, if someone knows, please comment it, I'll edit it.

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thanks I used mitra advice and got the same part as when you got to (sin+cos)/(sin^2+cos^2) but I thought this would end up making it 1/sin +1/cos which is csc+sec but I guess not. –  James Jul 16 '12 at 21:50
1  
@James: $(\sin + \cos)/(\sin^2+\cos^2)$ lets you use $\sin^2 + \cos^2 = 1$ on the bottom and you're done. Not sure how you got $1/\sin + 1/\cos$ but it's definitely an error. Perhaps you thought $\sin^2 + \cos^2 = (\sin + \cos)^2$? This would lead to $1/(\sin+\cos)$ and a second error would lead you to $1/\sin + 1/\cos$ perhaps. –  Fixee Jul 17 '12 at 14:29

For the first problem, use $\sec A = \frac{1}{\cos A}$, and $\csc A = \frac{1}{\sin A}$, then put them over a common denominator. Then convert $\tan$ and $\cot$ to $\sin$ and $\cos$ as you have done and put them over the same common denominator. Cancel these denominators and use $\sin^2 A + \cos^2 A = 1$ and you're done.

For the second one, begin as you did with $\frac{\cos}{\sin} + \frac{\sin}{1+\cos}$ and put them each over a denominator of $\sin^2$. Hint: You can convert $1+\cos$ to $\sin^2$ by multiplying it by $1-\cos$ and using the fact that $1-\cos^2=\sin^2$. The rest is just algebra.

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