# Proving this trig identity:$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$

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

What I've tried,

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

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

After simplifying,

$$=\frac{2(1+\sin\theta \cos\theta + \sin\theta + \cos\theta)}{\cos\theta(\cos\theta+2)}$$

I cant carry on further. Thus, any kind assistance would be much appreciated.

• – lab bhattacharjee Sep 22 '15 at 8:58
• @Abhishekstudent Most probably you are in 10th standard – Display name Sep 22 '15 at 9:04
• Yes I am in 10th standard. Cbse – Abhishekstudent Sep 22 '15 at 9:21
• Yes I am @Displayname. CBSE – Abhishekstudent Sep 22 '15 at 9:22

Multiplying and dividing by $(1+\sin\theta -\cos \theta)$ yields

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

Substituting $-\cos^2 \theta = \sin^2 \theta -1$, we obtain

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

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

Edit: clarifications for OP.

Consider the initial expression multiplied and divided by $(1+\sin\theta -\cos \theta)$:

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

Considering the terms grouped this way and using the fact that $(a+b)(a-b)=a^2-b^2$ we can conclude that

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

Expanding the square in the denominator leads to the LHS of the first equality.

• but $(a+b-c)$$^2=(a^2+b^2+c^2+2ab-2bc-2ac) Can u explain once please? @bharb – Abhishekstudent Sep 22 '15 at 9:38 • @Abhishekstudent : in the numerator we have (a+b+c)(a+b-c)=(a+b)^2-c^2, in the denominator (a+(b+c)(a-(b+c))=a^2-(b+c)^2. Is it clear? – Lonidard Sep 22 '15 at 9:42 • It should be \frac{1+cosθ+sinθ}{1+cosθ−sinθ}$$\frac{1+cos\theta-sin\theta}{1+cos\theta-sin \theta}$ So, the denominator is (a+b−c)2=(a2+b2+c2+2ab−2bc−2ac) @bharb – Abhishekstudent Sep 22 '15 at 10:52
• I'll extend the answer – Lonidard Sep 22 '15 at 10:53
• Yeah! Now I've got the answer! Thank you sir! Thank you very much!@bharb – Abhishekstudent Sep 22 '15 at 11:31

from Second last line, $$\displaystyle \frac{(1+\sin \theta+\cos \theta)^2}{(1+\cos \theta)^2-\sin^2 \theta}=\frac{2\left[1+\sin \theta +\sin \theta\cdot \cos \theta+\sin \theta\right]}{2\cos \theta \cdot (1+\cos \theta)} = \frac{2(1+\cos \theta)(1+\sin \theta)}{2\cos \theta\cdot (1+\cos \theta)}$$

So we get $$\displaystyle \frac{1+\sin \theta}{\cos \theta}$$

Here I have solved Using double angle formula.

Given $\displaystyle \bf{L.H.S}$ as $\displaystyle \frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$

Let $\theta=2\phi\;,$ Then $$\displaystyle \frac{1+\cos 2\phi+\sin 2\phi }{1+\cos 2\phi-\sin 2\phi} = \frac{2\cos^2 \phi+2\sin \phi\cdot \cos \phi}{2\cos^2 \phi-2\sin \phi\cdot \cos \phi}$$

Above we use the formula $$\bullet \; 1+\cos 2\phi = 2\cos^2 \phi$$ and $$\bullet\; 1-\cos 2\phi = 2\sin^2 \phi$$

and $$\bullet \; \sin 2\phi = 2\sin \phi\cdot \cos \phi$$ and $$\bullet\; \cos^2\phi-\sin^2 \phi = \cos 2\phi$$

So we get $$\displaystyle \frac{\cos \phi+\sin \phi}{\cos \phi-\sin \phi} = \frac{\cos \phi+\sin \phi}{\cos \phi-\sin \phi}\times \frac{\cos \phi+\sin \phi}{\cos \phi+\sin \phi} = \frac{\sin^2 \phi +\cos^2 \phi+\sin 2\phi}{\cos^2 \phi-\sin^2 \phi}$$

So we get $$\displaystyle \frac{1+\sin 2\phi}{\cos 2\phi}$$

Now Put $2\phi = \theta\;,$ We get

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

The denominator should be $2\cos\theta(1+\cos\theta)$

The numerator $=2(1+\sin\theta)(1+\cos\theta)$

• Can you explain it – Display name Sep 22 '15 at 9:15
• @Displayname, $$(1+\cos x)^2-\sin^2x=2\cos x+\cos^2x+1-\sin^2x=2\cos x+2\cos^2x=?$$ – lab bhattacharjee Sep 22 '15 at 9:20
• Ok got it, you were pointing out the mistake. +1 – Display name Sep 22 '15 at 9:27
• How? Can you please explain? – Abhishekstudent Sep 22 '15 at 11:41
• @Abhishekstudent, Please pinpoint your confusion – lab bhattacharjee Sep 22 '15 at 11:58

If you want to learn a more systematic approach to solve these problems you can read about Fourier Transforms which basically is about writing a function as sum of sines and cosines $$f(t) = \sum_{\forall k} a_k\cos(k\omega t) + b_k\sin(k\omega t)$$
Rewrite the equation so you get products instead of quotients, then these Fourier transforms have special rules which let you convolve them which automatically give you trigonometric one $$\sin^2(x) + \cos^2(x) = 1$$ double angle formulas like $$2\sin(x)\cos(x) = \sin(2x)$$ among many others. The functions will be very simple to convolve since their fourier transforms are very simple functions. $$\phantom{\mathcal{F}(lhs) = \mathcal{F}(rhs)}$$
You can try showing $$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta} -\frac{1+\sin\theta}{\cos\theta}=0$$ that is, $$(1+\cos\theta+\sin\theta)\cos\theta= (1+\cos\theta-\sin\theta)(1+\sin\theta)$$ Right-hand side: $$\def\ct#1{\cos^{#1}\theta}\def\st#1{\sin^{#1}\theta} (1-\st{})(1+\st{})+\ct{}(1+\st{})= 1-\st{2}+\ct{}(1+\st{})\\ =\ct{2}+\ct{}(1+\st{})=\ct{}(\ct{}+1+\st{})$$