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

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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

#1
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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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#1
+91510
+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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Very impressive, Melody  !!!

CPhill  Jan 26, 2015
#3
+91510
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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
+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
+91510
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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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