If `theta` lies in the first quadrant and `costheta=8/(17)` , then prove that `cos(pi/6+theta)+cos(pi/4-theta)+cos((2pi)/3-theta)=((sqrt(3)-1)/2+1/(sqrt(2)))(23)/(17)`

Explanation

To prove the equation

\[ \cos\left(\frac{\pi}{6} + \theta\right) + \cos\left(\frac{\pi}{4} - \theta\right) + \cos\left(\frac{2\pi}{3} - \theta\right) = \left(\frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}}\right)\frac{23}{17} \]

given that \(\cos \theta = \frac{8}{17}\) and \(\theta\) lies in the first quadrant, we can follow these steps:

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