# Trigonometric identity: $\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$

I have to prove the following result :

$$\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$$

I tried converting $\tan\theta$ & $\cot\theta$ into $\cos\theta$ and $\sin\theta$. That led to a huge expression which I wasn't able to simplify.

• your approach is right Jun 25, 2013 at 13:02
• possible duplicate of Another trigonometric proof...? Jun 26, 2013 at 5:33

You are on the right track.

writing $\tan\theta$ as$\dfrac {\sin\theta}{\cos\theta}$ and $\cot\theta$ as $\dfrac {\cos\theta}{\sin\theta}$, we get

$\dfrac {\frac {\sin\theta}{\cos\theta} }{1-\frac {\cos\theta}{\sin\theta} }+\dfrac {\frac {\cos\theta}{\sin\theta} }{1-\frac {\sin\theta}{\cos\theta} }$

$= \dfrac {\sin^2\theta}{cos\theta\cdot(\sin\theta-\cos\theta)} + \dfrac {\cos^2\theta}{\sin\theta\cdot(\cos\theta-\sin\theta)}$ (how?)

$= \dfrac {\sin^2\theta}{\cos\theta\cdot(\sin\theta-\cos\theta)} - \dfrac {\cos^2\theta}{\sin\theta\cdot(\sin\theta-\cos\theta)}$

$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^2\theta}{\cos\theta}-\dfrac {\cos^2\theta}{\sin\theta})$

$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^3\theta-\cos^3\theta}{\sin\theta\cdot\cos\theta})$

$=\dfrac{\sin\theta-\cos\theta}{\sin\theta-\cos\theta}\dfrac{\big(\sin^2\theta+\sin\theta\cdot\cos\theta+\cos^2\theta)}{\sin\theta\cdot\cos\theta}$(how?)

$=1\cdot \dfrac{1+\sin\theta\cdot\cos\theta}{\sin\theta\cdot\cos\theta}$ (why?)

which is

$1+\sec\theta\cdot\csc\theta$

QED.

• thank you very much! I got stuck in the 5th step. could'nt simplify further
– user76849
Jun 25, 2013 at 13:04
• Practice more and you'll be fine. Jun 25, 2013 at 13:05
• Use of the TeX commands \sin,\cos,\tan,\cot,\sec,\csc will make your answer look more appealing. They render as $\sin,\cos,\tan,\cot,\sec,\csc$. Jun 25, 2013 at 13:10
• Hah, @Lord_Farin, i just edited that for him. Jun 25, 2013 at 13:11
• @ThomasAndrews Commendable (noticing that you do this much more often)! I find it helps to leave an accompanying comment, so as to reduce (or at least, limit) the amount of future work. :) Jun 25, 2013 at 13:14

$$\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$$

$$=\frac{\tan^2\theta}{\tan\theta-1}+\frac{\cot\theta}{1-\tan\theta}(\text{ multiplying the first term by }\tan\theta )$$

$$=-\frac{\tan^2\theta}{1-\tan\theta}+\frac{\cot\theta}{1-\tan\theta}$$

$$=\frac{\cot\theta-\tan^2\theta}{1-\tan\theta}$$

$$=\frac{1-\tan^3\theta}{\tan\theta(1-\tan\theta)}$$

$$=\frac{1+\tan\theta+\tan^2\theta}{\tan\theta}(\text{ assuming }1-\tan\theta\ne0)$$

$$=1+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$$

$$=1+ \frac1{\sin\theta\cos\theta}$$

• that's a nice way too! Thank you very much..
– user76849
Jun 25, 2013 at 13:07
• Your approach is interesting & amazing. Could you please tell what made you think in the 1st step to multiply first term by tan theta & 2nd is , why did you not think to go for solving in sin,cos therms but instead tan. Aug 14, 2021 at 14:38

Just check once again. To begin with, it is correct to put $$tan(\theta) = s/c$$

( Using obvious abbreviations)

$$\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta} =\frac{s/c}{1-c/s}+\frac{c/s}{1-s/c}$$ $$=\frac{s^2/c -c^2/s}{s-c} =\frac{s^3 - c^3}{s.c.(s-c)}=\frac{s^2+c^2+s.c}{s.c}$$

$$=\frac{1+s.c}{s.c}=1+ 1/{(s.c)} = 1 + {\sec\theta}.{\csc\theta}$$

• How did you solve your 2nd step ? Aug 14, 2021 at 14:59
• as above, hope clear. Aug 14, 2021 at 15:37