Question

Find the values of the following.
(i) cos 38$°$ cos 52$°$ $-$ sin 38$°$ sin 52$°$

(ii) $\frac{\cos 80°}{\sin 10°}+\cos 59°\cos ec31°$

(iii) $\frac{2\tan 53°}{cot37°}-\frac{\cos 80°}{\tan 10°}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\cos 38°\cos 52°-\sin 38°\sin 52° \\ =\cos 38°\cos (90-38)°-\sin 38°\sin (90-38)° \\ =\cos 38°\sin 38°-\sin 38°\cos 38°[\because \cos (90-x)=\sin \mathit{}x,\sin (90-x)=\cos \mathit{}x] \\ =0 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{\cos 80°}{\sin 10°}+\cos 59°\mathrm{cosec}31° \\ =\frac{\cos (90-10)°}{\sin 10°}+\cos (90-31)°\mathrm{cosec}31° \\ =\frac{\sin 10°}{\sin 10°}+\sin 31°\mathrm{cosec}31°[\because \cos (90-x)=\sin x] \\ =1+1=2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathbf{Disclaimer}:\mathrm{It}\mathrm{would}\mathrm{be}\frac{2\tan 53°}{\cot 37°}-\frac{\cot 80°}{\tan 10°}\mathrm{in}\mathrm{stead}\mathrm{of}\frac{2\tan 53°}{\cot 37°}-\frac{\cos 80°}{\tan 10°} \\ \frac{2\tan 53°}{\cot 37°}-\frac{\cot 80°}{\tan 10°} \\ =\frac{2\tan (90-37)°}{\cot 37°}-\frac{\cot (90-10)°}{\tan 10°} \\ =\frac{2\cot 37°}{\cot 37°}-\frac{\tan 10°}{\tan 10°}[\because \tan (90-x)=\cot x,\cot (90-x)=\tan x] \\ =2\times 1-1=1 \end{gathered}$$