3
$\begingroup$

I need help with this one $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha $$

I tried moving sin a on the other side of the eqation $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0 $$ This are the operations I was able to do $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha*\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha - \sin \alpha = 0 $$

I don't see what else I can do with this, so I tried to solve the left part of the equation.

$$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \tan^2\alpha} - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - cos\alpha} + \frac{\sin\alpha\cdot\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha = \sin \alpha $$

And I get to nowhere again. I have no other ideas, I didn't see some formula or something. Any help is appreciated.

$\endgroup$
1
$\begingroup$

$$LHS=\frac{\sin^2 \alpha}{\sin \alpha-\cos \alpha}+\frac{\sin\alpha+\cos\alpha}{1-\tan^2\alpha}-\cos\alpha\\=\frac{\sin^2 \alpha}{\sin \alpha-\cos \alpha}-\frac{\cos^2\alpha}{\sin\alpha-\cos\alpha}-\cos\alpha\\=\sin\alpha+\cos\alpha-\cos\alpha=\sin\alpha$$

$\endgroup$
  • $\begingroup$ The quicker ways +1. $\endgroup$ – Mann May 17 '15 at 9:36
0
$\begingroup$

Hint: Put the fractions over the same denominator $\left(\sin\left(\alpha\right)-\cos\left(\alpha\right)\right)\left(1-\tan^{2}\left(\alpha\right)\right)$, then eliminate the denominator on the left side. Now you have just to expand the left and right side, and finally use the identity $\tan\left(\alpha\right)=\frac{\sin\left(\alpha\right)}{\cos\left(\alpha\right)}.$ You will get the same expression.

$\endgroup$
0
$\begingroup$

$\left(\dfrac{\sin^2 \alpha}{\sin \alpha+\cos\alpha}-(\sin\alpha+\cos\alpha)\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\left(\dfrac{\sin^2 \alpha-\sin^2 \alpha+\cos^2\alpha}{\sin \alpha-\cos\alpha}\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha}{\sin \alpha-\cos\alpha}+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha-\sin^2\alpha+\sin^2\alpha-\cos^2\alpha}{(\sin \alpha-\cos\alpha)(1-\tan^2\alpha)}\\=0$

$\endgroup$

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.