16k views

### Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?

Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$? It is said that this is derived from trigonometry, but I couldn't find why this is the case.
1k views

### What is the value of $\arctan(1/2)+\arctan(1/5)+\arctan(1/8)$?

What is the value of : $$\arctan(1/2)+\arctan(1/5)+\arctan(1/8)?$$ I tried to do geometric solution:: Where in the angles we are looking for are shown, but I can't solve it. Can we use it with ...
635 views

Find the values of $x$ such that $$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$. Checking for $x\in [-1,1]$ In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)... 2answers 3k views ### Derive the conditions$xy<1$for$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$and$xy>-1$for$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \text{, }xy<1\\ \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \text{, }xy>-1$$ But, How do I reach the conditions$xy<1$for the first ... 2answers 830 views ### Find$\tan x$if$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$Find$\tan x$if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$ First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$ So $$\... 4answers 226 views ### What is the value of \arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)? I was playing about with some numbers when I came up with this fun question. What is the value of \arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)? Here is my method: As is ... 2answers 185 views ### How do I find a Taylor series of \arctan \frac{2-2x}{1+4x} at x=0 I've been trying to find a Taylor series of \arctan \frac{2-2x}{1+4x} at x=0 The only thing I could think of was trying to find formula for the nth derivative but was unable to find it so it ... 4answers 139 views ### If \tan A=-\frac 1 2 and \tan B =-\frac 1 3, then A+B =? Answer should be in radians Like π/4 (45°) π(90°). I used \tan(A+B) formula and got 5/7 as the answer, but that's obviously wrong. 2answers 92 views ### Find all x such that \sin x = \frac{4}{5} and \cos x = \frac{3}{5}. Let$$ \left\{ \begin{array}{c} \sin x = \frac{4}{5} \\ \cos x = \frac{3}{5} \end{array} \right. $$Find all of the possible values for x. My try: By dividing the equations we obtain \... 3answers 303 views ### If \alpha = 2\arctan(2\sqrt{2}-1) and \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;, Then prove that \alpha>\beta If \alpha = 2\arctan(2\sqrt{2}-1) and \displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;, Then prove that \alpha>\beta \bf{My\; Try::} Given$$ \alpha ... 4answers 1k views ### Solve for$x$:$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)\$
$$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$ $$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$ $$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$ $$6x^2 + 5x -1 = 0$$ $$(6x-1)(x+1)=0$$ ...