If θ and Φ lie in the first quadrant such that

Given $\sin \theta=\frac{8}{17}$ and $\cos \phi=\frac{12}{13}$

$\cos \theta=\sqrt{\left(1-\sin ^{2} \theta\right)} \Rightarrow \sqrt{\left(1-\left(\frac{8}{17}\right)^{2}\right.}=\sqrt{\left(\frac{289-84}{289}\right)} \Rightarrow \sqrt{\left(\frac{225}{289}\right)}=\frac{15}{17}$

$\sin \phi=\sqrt{\left(1-\left(\frac{12}{13}\right)^{2}\right)} \Rightarrow \sqrt{\left(\frac{169-144}{169}\right)}=\sqrt{\left(\frac{25}{169}\right)} \Rightarrow \frac{5}{13}$

(i) $\sin (\theta-\Phi)=\sin \theta \cos \Phi+\cos \theta \sin \Phi$

$=\frac{8}{17} \cdot \frac{12}{13}+\frac{15}{17} \cdot \frac{5}{13} \Rightarrow \frac{96+75}{221}=\frac{171}{221}$

(ii) $\cos (\theta-\Phi)=\cos \theta \cdot \cos \Phi+\sin \theta \cdot \sin \Phi$

$=\frac{15}{17} \cdot \frac{12}{13}+\frac{8}{17} \cdot \frac{5}{13} \Rightarrow \frac{180+40}{221}=\frac{220}{221}$

(iii) We will first find out the Values of $\tan \theta$ and $\tan \Phi$,

$\tan \theta=\frac{\sin \theta}{\cos \theta} \Rightarrow \frac{8 / 17}{15 / 17}=\frac{8}{15} \operatorname{antan} \phi=\frac{\sin \phi}{\cos \phi} \Rightarrow \frac{5 / 13}{12 / 13}=\frac{5}{12}$+

$\tan (\theta-\Phi)=\tan (\theta-\phi)=\frac{\tan \theta-\tan \phi}{1+\tan \theta \cdot \tan \phi} \Rightarrow \frac{\frac{8}{15}-\frac{5}{12}}{1+\frac{8}{15} \cdot \frac{5}{12}}$