We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$

I have tried and it gets confusing.

  • $\begingroup$ @Mahesh The LHS isn't well defined. The denominator is $0\,$ $\leftarrow$ this comment was pre question edit. $\endgroup$ – Git Gud Jun 22 '13 at 15:55
  • $\begingroup$ Write the right hand side in terms of $\sin$ and $\cos$ and cross multiply. $\endgroup$ – Maesumi Jun 22 '13 at 16:00
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – Lord_Farin Jun 22 '13 at 16:47
  • $\begingroup$ I don't have any strong feelings if this is open or not; that I reopened it is basically an artificat of a quirk of the system. $\endgroup$ – quid Mar 4 at 20:18


$$=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}(\text{ dividing the numerator & the denominator by} \cos\theta )$$

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




Alternatively using Double-angle formula by putting $\tan\frac\theta2=t,$

$$\text{ LHS= }\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}+1}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-1}$$

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

$$\text{ RHS= }\frac1{\sec\theta-\tan\theta}=\frac1{\frac{1+t^2}{1-t^2}-\frac{2t}{1-t^2}}=\frac{1-t^2}{(1-t)^2}=\frac{1+t}{1-t} \text{assuming }t-1\ne0$$

  • $\begingroup$ @Ovi, thanks. Hope I could clear the idea. $\endgroup$ – lab bhattacharjee Jun 22 '13 at 16:27
  • $\begingroup$ Great! Simplest of all the answers $\endgroup$ – Mahesh Natarajan Jun 23 '13 at 6:04
  • $\begingroup$ @MaheshNatarajan, thanks. How about my other answer? $\endgroup$ – lab bhattacharjee Jun 23 '13 at 6:07

$$\frac{\sin\theta -\cos\theta +1}{\sin\theta +\cos\theta -1}= \frac{1}{\sec\theta - \tan\theta}$$

By taking $$\mbox{L.H.S ( Left hand side )} = \frac{\sin\theta -\cos\theta +1}{\sin\theta +\cos\theta -1} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}$$ [By applying $1-\cos\theta = 2\sin^2\frac{\theta}{2}$]

Which after simplification gives : $$\frac{\sin\frac{\theta}{2}+ \cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}$$

Now taking R.H.S we get : $$ \begin{align} \frac{1}{\sec\theta - \tan\theta} &= \frac{\cos\theta}{1-\sin\theta}\\ &= \frac{\cos^2\theta - \sin^2\theta}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \\ &= \frac{(\cos^2\frac{\theta}{2}-\sin^\frac{\theta}{2})}{(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})^2} \\ &= \frac{\sin\frac{\theta}{2}+ \cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}\\ &= \mbox{L.H.S ( Left hand side )}\end{align},$$ where the second equality comes from applying $\cos\theta = \cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$ and $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$.

  • $\begingroup$ Nice work, sultan +1. (Hint: when formatting trig functions, use \<trig-function like \sin\theta to get $\sin \theta$ instead of sin\theta, where you get $sin\theta$. $\endgroup$ – Namaste Jun 22 '13 at 16:52
  • $\begingroup$ @amWhy, just thought this method and added it to my answer, then found your solution. Needed to remove it which did cost me two revisions. Anyway, good work $+1$ $\endgroup$ – lab bhattacharjee Jun 22 '13 at 19:01

Let $\displaystyle{\sin\theta=\frac{a}{c}}$ and $\displaystyle{\cos\theta=\frac{b}{c}}$ such that $\displaystyle{a^2 + b^2=c^2}$.

$$\displaystyle{\frac{sin\theta - \cos\theta + 1}{\sin\theta+\cos\theta -1}}$$ $$\displaystyle{=\frac{a-b+c}{a+b-c}}$$

Now $\displaystyle{\tan\theta=\frac{a}{b}}$ and $\displaystyle{\sec\theta=\frac{c}{b}}$. Hence, $$\displaystyle{\frac{1}{\sec\theta-\tan\theta}}$$ $$\displaystyle{=\frac{1}{\frac{c}{b}-\frac{a}{b}}}$$ $$\displaystyle{=\frac{b}{c-a}}$$

Let us assume that $$\displaystyle{=\frac{a-b+c}{a+b-c}\neq\frac{b}{c-a}}$$

This simplifies to $$\displaystyle{a^2 + b^2\neq c^2}$$

This is obviously false. Hence, $$\displaystyle{\frac{a-b+c}{a+b-c}=\frac{b}{c-a}}$$


  • $\begingroup$ Aren't you proving an identity by first assuming the identity is true? $\endgroup$ – DJohnM Jun 22 '13 at 18:17
  • $\begingroup$ @User58220- Yes. But on assuming this, I've come upon an expression that cannot be refuted. Had I made an erroneous assumption, I would never have arrived upon a mathematically true expression. But if this is a little difficult to digest as of now, we can assume $\displaystyle{\frac{a-b+c}{a+b-c}\neq\frac{b}{c-a}}$. Then we'll get $\displaystyle{a^2 + b^2\neq c^2}$, which is a contradiction. $\endgroup$ – fierydemon Jun 23 '13 at 1:47
  • $\begingroup$ Why not taking $c=1$? $\endgroup$ – Taladris Jun 23 '13 at 3:50
  • $\begingroup$ @Taladris- yes you can. My proof is just a little more general, where $c$ can $\neq 1$ $\endgroup$ – fierydemon Jun 23 '13 at 4:24
  • $\begingroup$ @AyushKhaitan Your contradiction proof is right, but your other statements are wrong. "If $P$, then $Q$" does not prove $P$ just because $Q$ is true. (Think, for example, of multiplying both sides of $1=2$ by $0$.) $\endgroup$ – Alexander Gruber Jun 23 '13 at 15:19

These kind of questions often benefit from the identity $a^2-b^2=(a-b)(a+b)$ in conjunction with Pythagorean trig identitities. Here,

$$ \begin{align} \frac{\sin t-\cos t+1}{\sin t+\cos t -1}&=\frac{(\sin t+1)-\cos t}{(\sin t+\cos t) -1}\cdot\frac{(\sin t+1)+\cos t}{(\sin t+\cos t) +1}\\ &=\frac{\sin^2 t+2\sin t+1-\cos^2 t}{\sin^2 t+2\sin t\cos t+\cos^2 t -1}\\ &=\frac{2\sin^2 t+2\sin t}{2\sin t\cos t}\\ &=\frac{\sin t+1}{\cos t}\\ &=\tan t+\sec t\\ &=(\tan t+\sec t)\frac{\tan t-\sec t}{\tan t-\sec t}\\ &=\frac{\tan^2 t-\sec^2 t}{\tan t-\sec t}\\ &=\frac{1}{\sec t-\tan t}\\ \end{align} $$


multiply numerator and denominator by $(1-\sin\theta)$. $$\dfrac {\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\times \dfrac{1-\sin\theta}{1-\sin\theta}$$ $$\dfrac {\sin\theta-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta+1-\sin\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$ $$\dfrac {1-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$ $$\dfrac {\cos^2\theta-\cos\theta+\sin\theta\cdot\cos\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$ $$\dfrac {\cos\theta(\cos\theta+\sin\theta-1)}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$ $$\dfrac {\cos\theta}{1-\sin\theta}$$ $$\dfrac {1}{\dfrac{1-\sin\theta}{\cos\theta}}$$ $$\dfrac {1}{\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{{\cos\theta}}}$$ $$\dfrac {1}{\sec\theta-\tan\theta}$$


$1.$ One of the ways to prove Trigonometric identities involves aiming at the Right Hand Side(RHS),

In the first method of my other answer, I have eliminated $\sin\theta,\cos\theta$ (which are absent in the RHS) to introduce $\sec\theta,\tan\theta$

As the RHS $(=\frac1{\sec\theta-\tan\theta}=\frac{\cos\theta}{1-\sin\theta})$ involves $1-\sin\theta,$ iostream007 in his answer, has introduced $1-\sin\theta$ and kept it as is in the denominator and manipulated it in the numerator

As the RHS involves $\cos\theta,$ I will here multiply the numerator & the denominator of the LHS by $\cos\theta$ and leave $\cos\theta$ unaffected in the numerator.

The denominator$\cdot \cos\theta$

$$=(\sin\theta + \cos\theta - 1)\cos\theta=\sin\theta\cos\theta + \cos^2\theta - \cos\theta$$ $$=\sin\theta\cos\theta + 1-\sin^2\theta - \cos\theta=(1-\sin\theta)(1+\sin\theta)-\cos\theta(1-\sin\theta)=(1-\sin\theta)(1+\sin\theta-\cos\theta)$$

$$\implies \dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1}=\dfrac{(\sin\theta - \cos\theta + 1)\cos\theta}{(\sin\theta + \cos\theta - 1)\cos\theta}$$

$$=\dfrac{(\sin\theta - \cos\theta + 1)\cos\theta}{(1-\sin\theta)(1+\sin\theta-\cos\theta)}=\dfrac{\cos\theta}{1-\sin\theta}$$

$2.$ Another way is to make both sides reaching at the same expression

This approach has been used in sultan's answer as well as the second method of my other answer

  • $\begingroup$ @MaheshNatarajan, the $3$rd approach is to rationalize the denominator (multiplying by something like conjugate) used by alex.jordan $\endgroup$ – lab bhattacharjee Jun 23 '13 at 6:09

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