+0  
 
0
1
1233
5
avatar

Prove that tan(a)/(1-cot(a))+cot(a)/(1-tan(a))= sec(A)*csc(A)+1

Guest Jan 25, 2015

Best Answer 

 #1
avatar+92805 
+10

Prove that tan(a)/(1-cot(a))+cot(a)/(1-tan(a))= sec(A)*csc(A)+1

 

$$\\\frac{tan(a)}{(1-cot(a))}+\frac{cot(a)}{(1-tan(a))}= sec(A)*csc(A)+1\\\\
LHS=\frac{\frac{sina}{cosa}}{(1-\frac{cosa}{sina})}+\frac{\frac{cosa}{sina}}{(1-\frac{sina}{cosa})}\\\\
LHS=\frac{\frac{sina}{cosa}}{(\frac{sina-cosa}{sina})}+\frac{\frac{cosa}{sina}}{(\frac{cosa-sina}{cosa})}\\\\
LHS=\frac{sina}{cosa}\div (\frac{sina-cosa}{sina}) +\frac{cosa}{sina}\div(\frac{cosa-sina}{cosa}) \\\\
LHS=\frac{sina}{cosa}\times(\frac{sina}{sina-cosa}) +\frac{cosa}{sina}\times(\frac{cosa}{cosa-sina}) \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} +\frac{cos^2a}{sina(cosa-sina)} \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} -\frac{cos^2a}{sina(sina-cosa)} \\\\
LHS=\frac{sin^3a}{cosa*sina(sina-cosa)} -\frac{cos^3a}{cosa*sina(sina-cosa)} \\\\$$

 

$$\\LHS=\frac{sin^3a-cos^3a}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sina-cosa)(sin^2a+sinacosa+cos^2a)}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sinacosa+cos^2a+sin^2a)}{cosa*sina} \\\\
LHS=\frac{(sinacosa+1)}{cosa*sina} \\\\
LHS=1+\frac{1}{cosa*sina} \\\\
LHS=sec(a)*cosec(a)+1 \\\\
LHS=RHS\qquad QED \\\\$$

Melody  Jan 26, 2015
 #1
avatar+92805 
+10
Best Answer

Prove that tan(a)/(1-cot(a))+cot(a)/(1-tan(a))= sec(A)*csc(A)+1

 

$$\\\frac{tan(a)}{(1-cot(a))}+\frac{cot(a)}{(1-tan(a))}= sec(A)*csc(A)+1\\\\
LHS=\frac{\frac{sina}{cosa}}{(1-\frac{cosa}{sina})}+\frac{\frac{cosa}{sina}}{(1-\frac{sina}{cosa})}\\\\
LHS=\frac{\frac{sina}{cosa}}{(\frac{sina-cosa}{sina})}+\frac{\frac{cosa}{sina}}{(\frac{cosa-sina}{cosa})}\\\\
LHS=\frac{sina}{cosa}\div (\frac{sina-cosa}{sina}) +\frac{cosa}{sina}\div(\frac{cosa-sina}{cosa}) \\\\
LHS=\frac{sina}{cosa}\times(\frac{sina}{sina-cosa}) +\frac{cosa}{sina}\times(\frac{cosa}{cosa-sina}) \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} +\frac{cos^2a}{sina(cosa-sina)} \\\\
LHS=\frac{sin^2a}{cosa(sina-cosa)} -\frac{cos^2a}{sina(sina-cosa)} \\\\
LHS=\frac{sin^3a}{cosa*sina(sina-cosa)} -\frac{cos^3a}{cosa*sina(sina-cosa)} \\\\$$

 

$$\\LHS=\frac{sin^3a-cos^3a}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sina-cosa)(sin^2a+sinacosa+cos^2a)}{cosa*sina(sina-cosa)} \\\\
LHS=\frac{(sinacosa+cos^2a+sin^2a)}{cosa*sina} \\\\
LHS=\frac{(sinacosa+1)}{cosa*sina} \\\\
LHS=1+\frac{1}{cosa*sina} \\\\
LHS=sec(a)*cosec(a)+1 \\\\
LHS=RHS\qquad QED \\\\$$

Melody  Jan 26, 2015
 #2
avatar+87303 
0

Very impressive, Melody  !!!

 

CPhill  Jan 26, 2015
 #3
avatar+92805 
0

Thanks Chris,

I am waiting for Heureka or Alan to come along and do it in 3 or 4 lines  LOL   

Melody  Jan 26, 2015
 #4
avatar+1037 
+5

Another Solution

 

$$Prove \\\\
\frac{\tan \left(a\right)}{1-\cot \left(a\right)}+\frac{\cot \left(a\right)}{1-\tan \left(a\right)}=\sec \left(a\right)\csc \left(a\right)+1 \\$
\\
LHS
\\\\
\hspace*{0.7cm}\ $=\frac{-\tan ^2\left(a\right)+\tan \left(a\right)-\cot ^2\left(a\right)+\cot \left(a\right)}{\left(\tan \left(a\right)-1\right)\left(\cot \left(a\right)-1\right)}$ \\

$=\frac{-\left(\frac{\cos \left(a\right)}{\sin \left(a\right)}\right)^2-\left(\frac{\sin \left(a\right)}{\cos \left(a\right)}\right)^2+\frac{\cos \left(a\right)}{\sin \left(a\right)}+\frac{\sin \left(a\right)}{\cos \left(a\right)}}{\left(-1+\frac{\cos \left(a\right)}{\sin \left(a\right)}\right)\left(-1+\frac{\sin \left(a\right)}{\cos \left(a\right)}\right)} \\\\

$=\frac{\cos ^2\left(a\right)+\sin ^2\left(a\right)+\cos \left(a\right)\sin \left(a\right)}{\cos \left(a\right)\sin \left(a\right)}\\\

\hspace*{3.0cm}$\mathrm{Using \; identity}: \cos ^2\left(x\right)+\sin ^2\left(x\right)=1 \\$

$=\frac{1+\cos \left(a\right)\sin \left(a\right)}{\sin \left(a\right)\cos \left(a\right)}$
\\\\$$

 

$$RHS \\\\
\sec \left(a\right)\csc \left(a\right)+1 \\$

\ Transform using Sin and Cos \\

$=1+\frac{1}{\cos \left(a\right)}\frac{1}{\sin \left(a\right)}\\\

\ Simplify \\

$=\frac{1+\cos \left(a\right)\sin \left(a\right)}{\cos \left(a\right)\sin \left(a\right)}\\$$

 

Snarky Comments pending . . ..

Nauseated  Jan 26, 2015
 #5
avatar+92805 
0

This answer is quite similar to mine Nauseated.

You may have made it a smidge easier by getting a common denominator in the first place but otherwise our answers are the same.    

 

SEE great minds do think alike!  

Melody  Jan 26, 2015

13 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.