Precise Carving

Let $\alpha = \tan^{-1} \left( \frac 34 \right)$ , $\beta = \tan^{-1} \left( \frac 15 \right)$ and $\gamma = \cos^{-1} \left( \frac cd \right)$ . Therefore, we have:

$\begin{aligned} \alpha - 2 \beta & = \gamma \\ \implies \tan \gamma & = \tan (\alpha - 2 \beta) \\ & = \frac {\tan \alpha - \tan 2\beta}{1 + \tan \alpha \tan 2\beta} \\ & = \frac {\frac 34 - \frac {\frac 25}{1-\frac 1{25}}}{1 + \frac 34 \cdot \frac {\frac 25}{1-\frac 1{25}}} \\ & = \frac {\frac 34 - \frac 5{12}}{1 + \frac 34 \cdot \frac 5{12}} \\ & = \frac {16}{63} \end{aligned}$

Now, we have:

$\begin{aligned} \frac cd & = \cos \gamma = \frac 1{\sec \gamma} = \frac 1{\sqrt{\sec^2 \gamma}} = \frac 1{\sqrt{1+\tan^2 \gamma}} = \frac 1{\sqrt{1+ \left(\frac {16}{63} \right)^2}} = \frac {63}{65} \end{aligned}$

$\implies c + d = 63+65 = \boxed{128}$

Instead of applying the sum/difference formula, it is often easier to manipulate the complex number interpretation. This approach can be easily generalized.

First, observe that:

$\dfrac{4 + 3i}{(5 +i)^2} = \dfrac{4+3i}{24+10i} = \dfrac{(4+3i)(24-10i)}{(24-10i)(24+10i)} = \dfrac{126 + 32i}{24^2 + 10^2} = \dfrac{63}{338} + \dfrac8{169} i$

If we take the polar representation of this number, we get that:

$\frac{ \sqrt{4^2 + 3^2 } CIS ( \tan^{-1} \frac{3}{4} ) } { \left[ \sqrt{ 1^2 + 5^2 } CIS ( \tan^{-1} \frac {1}{5} ) \right]^2 } = \sqrt{ ( \frac{ 63}{338} ) ^2 + ( \frac{8}{169} ) ^2 } CIS \left( \cos ^{-1} \frac{ \frac{ 63}{338} } { \sqrt{ ( \frac{ 63}{338} ) ^2 + ( \frac{8}{169} ) ^2 } } \right)$

Now, simplifying the calculations and considering the angle measure, we get that:

$\tan^{-1} \frac{3}{4} - 2 \times \tan^{-1} \frac {1}{5} = \cos^{-1} \frac{63}{65}$