# Prove that

Question:

If $\sin x=\frac{12}{3}$ and $\sin y=\frac{4}{5}$ where $\frac{\pi}{2} (i)$\sin (x+y)$(ii)$\cos (x+y)$(iii)$\tan (x-y)$Solution: Given$\sin x=\frac{12}{13}$and$\sin y=\frac{4}{5}$, Here we will find values of cosx and cosy$\cos x=\sqrt{\left(1-\sin ^{2} x\right)} \Rightarrow \sqrt{\left(1-\left(\frac{12}{13}\right)^{2}\right)}=\sqrt{\left(\frac{169-144}{169}\right)} \Rightarrow \sqrt{\left(\frac{25}{169}\right)}=\frac{5}{13}\cos y=\sqrt{\left(1-\sin ^{2} y\right)} \Rightarrow \sqrt{\left(1-\left(\frac{4}{5}\right)^{2}\right)}=\sqrt{\left(\frac{25-16}{25}\right)} \Rightarrow \sqrt{\left(\frac{9}{25}\right)}=\frac{3}{5}$(i)$\sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y\Rightarrow \frac{12}{13} \cdot \frac{3}{5}+\frac{5}{13} \cdot \frac{4}{5} \Rightarrow \frac{36+20}{65}=\frac{56}{65}$(ii)$\cos (x+y)=\cos x \cdot \cos y+\sin x \cdot \sin y=\frac{5}{13} \cdot \frac{3}{5}+\frac{12}{13} \cdot \frac{4}{5} \Rightarrow \frac{15+48}{65}=\frac{63}{65}$(iii) Here first we will calculate value of tanx and tany,$\tan x=\frac{\sin x}{\cos x} \Rightarrow \frac{12 / 13}{5 / 13}=\frac{5}{12}$and tany$=\frac{\sin y}{\cos y} \Rightarrow \frac{4 / 5}{3 / 5}=\frac{4}{3}\tan (\mathrm{x}-\mathrm{y})=\frac{\tan \mathrm{x}-\operatorname{tany}}{1+\tan \mathrm{x} \cdot \tan \mathrm{y}} \Rightarrow \frac{\frac{5}{12}-\frac{4}{3}}{1+\frac{5}{12} \cdot \frac{4}{3}}=\frac{\frac{5-16}{12}}{\frac{36+20}{36}} \Rightarrow \frac{\frac{-11}{12}}{\frac{56}{36}}=\frac{-33}{56}\$