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The question asks to prove: $$ \frac{\tan{A}}{1-\cot{A}}+ \frac{\cot{A}}{1-\tan{A}}=\sec{A}\csc{A}+1$$ using only: $$ \sin^2{A}+\cos^2{A}=1\;\; \text{ & }\; \;\tan^2{A}+1=\sec^2{A}\;\; \text{ & } \;\; \cot^2{A}+1=\csc^2{A} $$


My solution is join the two fractions, then multiplying numerator and denominator by $\tan^2{A}$ to have only tangent functions: $$\frac{\tan{A}-\tan^2{A}+\cot{A}-\cot^2{A}}{ 2-\cot{A}-\tan{A} } \rightarrow \frac{\tan^3{A}-\tan^4{A}+\tan{A}-1}{2\tan^2{A}-\tan{A}-\tan^3{A}}$$ This then apparently factorises (I would never have spotted the top, it's only after trying it on Mathematica that I realised it did) to give:

$$ \frac{-(\tan{A}-1)^2(\tan^2{A}+\tan{A}+1)}{-\tan{A}(\tan{A}-1)^2} =\cot{A}(\tan^2{A}+\tan{A}+1)=1+\cot{A}+\tan{A}$$ Which then simplifies down to the required result, but needless to say this isn't very elegant and not inline with the previous problems that whilst not easy didn't involve factoring a quartic which leads me to believe there is certainly an easier way to do it hence the question, I just can't seem to see it (I've done too many for today!).

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I can't think of an "elegant" solution, but typically just converting everything to sin/cos is the way to go. For example, let $x=\sin(A)$ and $y=\cos(A)$. Then, $$ \frac{x/y}{1-y/x}+\frac{y/x}{1-x/y} = \frac{x^2/y}{x-y} - \frac{y^2/x}{x-y} = \frac{x^3-y^3}{xy(x-y)} = \frac{x^2+xy+y^2}{xy} = 1 + \frac{x^2+y^2}{xy}.$$

Now making use of the trig identity, we obtain simply $1+\frac{1}{xy}$, which is the result we want.

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  • $\begingroup$ This was the way I tried to do it at first, but I kept getting stuck where nothing seemed to simplify further (I bet it was some stupid notational mistake since it seems relatively straight forward as you've laid it out, much more inline with the previous questions) so I tried alternative ways. It's so much clearer using $x$'s and $y$'s I'll make use of it in future problems! Thanks $\endgroup$ – Jay Apr 4 '14 at 0:03
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Prove the following simpler identities first, and they help build a more elegant solution:

Identity 1 $$tanA+cotA=\frac{sinA}{cosA}+\frac{cosA}{sinA}=\frac{sin^2A+cos^2A}{sinAcosA}=secAcscA$$

Identity 2 $$sec^2A+csc^2A=\frac{1}{cos^2A}+\frac{1}{sin^2A}=\frac{sin^2A+cos^2A}{sin^2Acos^2A}=sec^2Acsc^2A$$

Identity 3 (using $cotA=\frac{1}{tanA}$ and Identity 1) $$(1-cotA)(1-tanA)=1-cotA-tanA+cotAtanA=2-cotA-tanA=2-secAcscA$$

Identity 4 (using Idenity 3) $$tan^2A+cot^2A=(sec^2A-1)+(csc^2A-1)=sec^2A+csc^2A-2=sec^2Acsc^2A-2$$

With these identities, the following simplification can be made: $$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=\frac{(tanA-tan^2A)+(cotA-cot^2A)}{(1-cotA)(1-tanA)}=\frac{tanA+cotA-(tan^2A+cot^2A)}{2-secAcscA}=\frac{2+secAcscA-sec^2Acsc^2A}{2-secAcscA}=\frac{(2-secAcscA)(1+secAcscA)}{2-secAcscA}=1+secAcscA$$

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