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

 Jan 25, 2015

Best Answer 

 #1
avatar+118703 
+10

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

 

tan(a)(1cot(a))+cot(a)(1tan(a))=sec(A)csc(A)+1LHS=sinacosa(1cosasina)+cosasina(1sinacosa)LHS=sinacosa(sinacosasina)+cosasina(cosasinacosa)LHS=sinacosa÷(sinacosasina)+cosasina÷(cosasinacosa)LHS=sinacosa×(sinasinacosa)+cosasina×(cosacosasina)LHS=sin2acosa(sinacosa)+cos2asina(cosasina)LHS=sin2acosa(sinacosa)cos2asina(sinacosa)LHS=sin3acosasina(sinacosa)cos3acosasina(sinacosa)

 

LHS=sin3acos3acosasina(sinacosa)LHS=(sinacosa)(sin2a+sinacosa+cos2a)cosasina(sinacosa)LHS=(sinacosa+cos2a+sin2a)cosasinaLHS=(sinacosa+1)cosasinaLHS=1+1cosasinaLHS=sec(a)cosec(a)+1LHS=RHSQED

 Jan 26, 2015
 #1
avatar+118703 
+10
Best Answer

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

 

tan(a)(1cot(a))+cot(a)(1tan(a))=sec(A)csc(A)+1LHS=sinacosa(1cosasina)+cosasina(1sinacosa)LHS=sinacosa(sinacosasina)+cosasina(cosasinacosa)LHS=sinacosa÷(sinacosasina)+cosasina÷(cosasinacosa)LHS=sinacosa×(sinasinacosa)+cosasina×(cosacosasina)LHS=sin2acosa(sinacosa)+cos2asina(cosasina)LHS=sin2acosa(sinacosa)cos2asina(sinacosa)LHS=sin3acosasina(sinacosa)cos3acosasina(sinacosa)

 

LHS=sin3acos3acosasina(sinacosa)LHS=(sinacosa)(sin2a+sinacosa+cos2a)cosasina(sinacosa)LHS=(sinacosa+cos2a+sin2a)cosasinaLHS=(sinacosa+1)cosasinaLHS=1+1cosasinaLHS=sec(a)cosec(a)+1LHS=RHSQED

Melody Jan 26, 2015
 #2
avatar+130477 
0

Very impressive, Melody  !!!

 

 Jan 26, 2015
 #3
avatar+118703 
0

Thanks Chris,

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

 Jan 26, 2015
 #4
avatar+1038 
+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 \\$    $=\frac{1+\cos \left(a\right)\sin \left(a\right)}{\sin \left(a\right)\cos \left(a\right)}$  \\\\

 

RHSsec(a)csc(a)+1$ TransformusingSinandCos$=1+1cos(a)1sin(a)  Simplify$=1+cos(a)sin(a)cos(a)sin(a)

 

Snarky Comments pending . . ..

 Jan 26, 2015
 #5
avatar+118703 
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!  

 Jan 26, 2015

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