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

Guest Jan 25, 2015

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 #1
avatar+92206 
+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
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5+0 Answers

 #1
avatar+92206 
+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+85726 
0

Very impressive, Melody  !!!

 

CPhill  Jan 26, 2015
 #3
avatar+92206 
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+92206 
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

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