Prove that


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) $\tan (x-y)$



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 $=\frac{\sin y}{\cos y} \Rightarrow \frac{4 / 5}{3 / 5}=\frac{4}{3}$

$\tan (\mathrm{x}-\mathrm{y})=\frac{\tan \mathrm{x}-\operatorname{tany}}{1+\tan \mathrm{x} \cdot \tan \mathrm{y}} \Rightarrow \frac{\frac{5}{12}-\frac{4}{3}}{1+\frac{5}{12} \cdot \frac{4}{3}}=\frac{\frac{5-16}{12}}{\frac{36+20}{36}} \Rightarrow \frac{\frac{-11}{12}}{\frac{56}{36}}=\frac{-33}{56}$


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