3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – David Mitra Jul 16 '12 at 21:34
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – James Jul 16 '12 at 21:50
  • 1
    $\begingroup$ @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. $\endgroup$ – Fixee Jul 17 '12 at 14:29
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.