1. ## Inverse tan

Using the expansion of $tan (a+b)$ find the value of $k$ such that $tan^{-1} k + tan^{-1} (\frac{3}{5}) = \frac{\pi}{4}$

2. Originally Posted by nerdzor
Using the expansion of $tan (a+b)$ find the value of $k$ such that $tan^{-1} k + tan^{-1} (\frac{3}{5}) = \frac{\pi}{4}$
Let $\tan^{-1} k = a \Rightarrow \tan a = k$ and let $\tan^{-1} \frac{3}{5} = b \Rightarrow \tan b = \frac{3}{5}$ and let [tex].

Then $a + b = \frac{\pi}{4} \Rightarrow \tan (a + b) = 1$.

Now substitute the expressions for $\tan a$ and $\tan b$ into the expansion of $\tan (a + b)$ and solve for $k$.

3. Hello, nerdzor!

We can do this head-on ... if you dare.

We need: . $\tan (A + B) \:=\:\frac{\tan A + \tan B}{1 - \tan A\tan B}$

Using the expansion of $\tan (A+B)$, find the value of $k$

such that: . $\tan^{-1}(k) + \tan^{-1}\!\left(\tfrac{3}{5}\right) \:=\:\frac{\pi}{4}$

Take the tangent of both sides: . $\tan\bigg[\tan^{-1}(k) + \tan^{-1}\!\left(\tfrac{3}{5}\right)\bigg] \;=\;\tan\left(\tfrac{\pi}{4}\right)$

. . . . $\frac{\tan\left[\tan^{-1}(k)\right] + \tan\left[\tan^{-1}(\frac{3}{5})\right]} {1 - \tan\left[\tan^{-1}(k)\right]\,\tan\left[\tan^{-1}(\frac{3}{5})\right]} \;=\;1$ . $\Longrightarrow\quad \frac{k + \frac{3}{5}}{1 - \frac{3}{5}k} \:=\:1$

. . . . $k + \frac{3}{5} \;=\;1 - \frac{3}{5}k \quad\Longrightarrow\quad \frac{8}{5}k \;=\;\frac{2}{5} \quad\Longrightarrow\quad \boxed{k \:=\:\frac{1}{4}}$