Question

Prove that:
(i) tan 225° cot 405° + tan 765° cot 675° = 0
(ii)
(iii) cos 24° + cos 55° + cos 125° + cos 204° + cos 300° =
(iv) tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
(v) cos 570° sin 510° + sin (−330°) cos (−390°) = 0
(vi)
(vii)

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=\tan 225°\cot 405°+\tan 765°\cot 675° \\ =\tan \left(90°\times 2+45°\right)\cot \left(90°\times 4+45°\right)+\tan \left(90°\times 8+45°\right)\cot \left(90°\times 7+45°\right) \\ =\tan \left(45°\right)\cot \left(45°\right)+\tan \left(45°\right)\left[-\tan \left(45°\right)\right] \\ =1\times 1+1\times \left(-1\right) \\ =1-1 \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=\sin \frac{8\mathrm{\pi }}{3}\cos \frac{23\mathrm{\pi }}{6}+\cos \frac{13\mathrm{\pi }}{3}\sin \frac{35\mathrm{\pi }}{6} \\ =\sin \left(\frac{8}{3}\times 180°\right)\cos \left(\frac{23}{6}\times 180°\right)+\cos \left({\displaystyle \frac{13}{3}}\times 180°\right)\sin \left({\displaystyle \frac{35}{6}}\times 180°\right) \\ =\sin \left(480°\right)\cos \left(690°\right)+\cos \left(780°\right)\sin \left(1050°\right) \\ =\sin \left(90°\times 5+30°\right)\cos \left(90°\times 7+60°\right)+\cos \left(90°\times 8+60°\right)\sin \left(90°\times 11+60°\right) \\ =\cos \left(30°\right)\sin \left(60°\right)+\cos \left(60°\right)\left[-\cos \left(60°\right)\right] \\ =\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \left(-\frac{1}{2}\right) \\ =\frac{3}{4}-\frac{1}{4} \\ =\frac{2}{4} \\ =\frac{1}{2} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{LHS}=\cos 24°+\cos 55°+\cos 125°+\cos 204°+\cos 300° \\ =\cos 24°+\cos \left(90°-35°\right)+\cos \left(90°\times 1+35°\right)+\cos \left(90°\times 2+24°\right)+\cos \left(90°\times 3+30°\right) \\ =\cos 24°+\sin 35°-\sin 35°-\cos 24°+\sin 30° \\ =0+0+\frac{1}{2} \\ =\frac{1}{2} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{LHS}=\tan \left(-225°\right)\cot \left(-405°\right)-\tan \left(-765°\right)\cot \left(675°\right) \\ =\left[-\tan \left(225°\right)\right]\left[-\cot \left(405°\right)\right]-\left[-\tan \left(765°\right)\right]\cot \left(675°\right)\left[\because \tan \left(-x\right)=\tan \left(x\right)\mathrm{and}\cot \left(-x\right)=-\cot \left(x\right)\right] \\ =\tan \left(225°\right)\cot \left(405°\right)+\tan \left(765°\right)\cot \left(675°\right) \\ =\tan \left(90°\times 2+45°\right)\cot \left(90°\times 4+45°\right)+\tan \left(90°\times 8+45°\right)\cot \left(90°\times 7+45°\right) \\ =\tan \left(45°\right)\cot \left(45°\right)+\tan \left(45°\right)\left[-\tan \left(45°\right)\right] \\ =1\times 1+1\times \left(-1\right) \\ =1-1 \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}=\cos \left(570°\right)\sin \left(510°\right)+\sin \left(-330°\right)\cos \left(-390°\right) \\ =\cos \left(570°\right)\sin \left(510°\right)+\left[-\sin \left(330°\right)\right]\cos \left(390°\right)\left[\because \sin \left(-x\right)=-\sin x\mathrm{and}\cos \left(-x\right)=\cos x\right] \\ =\cos \left(570°\right)\sin \left(510°\right)-\sin \left(330°\right)\cos \left(390°\right) \\ =\cos \left(90°\times 6+30°\right)\sin \left(90°\times 5+60°\right)-\sin \left(90°\times 3+60°\right)\cos \left(90°\times 4+30°\right) \\ =-\cos \left(30°\right)\cos \left(60°\right)-\left[-\cos \left(60°\right)\right]\cos \left(30°\right) \\ =-\cos \left(30°\right)\cos \left(60°\right)+\cos \left(30°\right)\sin \left(60°\right) \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{LHS}=\tan \frac{11\mathrm{\pi }}{3}-2\sin \frac{4\mathrm{\pi }}{6}-\frac{3}{4}\cos e{c}^2\frac{\mathrm{\pi }}{4}+4{\cos }^2\frac{17\mathrm{\pi }}{6} \\ =\tan \left({\displaystyle \frac{11\mathrm{\pi }}{3}}\right)-2\sin \left({\displaystyle \frac{4\mathrm{\pi }}{6}}\right)-\frac{3}{4}{\left[\cos ec\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)\right]}^2+4{\left[\cos \left({\displaystyle \frac{17\mathrm{\pi }}{6}}\right)\right]}^2 \\ =\tan \left({\displaystyle \frac{11}{3}}\times 180°\right)-2\sin \left({\displaystyle \frac{4}{6}}\times 180°\right)-\frac{3}{4}{\left[\cos ec\left({\displaystyle \frac{180°}{4}}\right)\right]}^2+4{\left[\cos \left({\displaystyle \frac{17\times 180°}{6}}\right)\right]}^2 \\ =\tan \left(660°\right)-2\sin \left(120°\right)-\frac{3}{4}{\left[\cos ec\left(45°\right)\right]}^2+4{\left[\cos \left(510°\right)\right]}^2 \\ =\tan \left(660°\right)-2\sin \left(120°\right)-\frac{3}{4}{\left[\mathrm{cosec}\left(45°\right)\right]}^2+4{\left[\cos \left(510°\right)\right]}^2 \\ =\tan \left(90°\times 7+30°\right)-2\sin \left(90°\times 1+30°\right)-\frac{3}{4}{\left[\mathrm{cosec}\left(45°\right)\right]}^2+4{\left[\cos \left(90°\times 5+60°\right)\right]}^2 \\ =\left[-\cot \left(30°\right)\right]-\left[2\cos \left(30°\right)\right]-\frac{3}{4}{\left[\mathrm{cosec}\left(45°\right)\right]}^2+4{\left[-\sin \left(60°\right)\right]}^2 \\ =-\cot \left(30°\right)-2\cos \left(30°\right)-\frac{3}{4}{\left[\mathrm{cosec}\left(45°\right)\right]}^2+4{\left[\sin \left(60°\right)\right]}^2 \\ =-\sqrt{3}-\frac{2\sqrt{3}}{2}-\frac{3}{4}{\left[\sqrt{2}\right]}^2+4{\left[\frac{\sqrt{3}}{2}\right]}^2 \\ =-\sqrt{3}-\sqrt{3}-\frac{3}{2}+3 \\ =\frac{3-4\sqrt{3}}{2} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(v\mathrm{ii}\right)\mathrm{LHS}=3\sin \frac{\mathrm{\pi }}{6}sec\frac{\mathrm{\pi }}{3}-4\sin \frac{5\mathrm{\pi }}{6}cot\frac{\mathrm{\pi }}{4} \\ =3\sin \left({\displaystyle \frac{180°}{6}}\right)\sec \left({\displaystyle \frac{180°}{3}}\right)-4\sin \left({\displaystyle \frac{5\times 180°}{6}}\right)\cot \left({\displaystyle \frac{180°}{4}}\right) \\ =3\sin \left(30°\right)\sec \left(60°\right)-4\sin \left(150°\right)\cot \left(45°\right) \\ =3\sin \left(30°\right)\sec \left(60°\right)-4\sin \left(90°\times 1+60°\right)\cot \left(45°\right) \\ =3\sin \left(30°\right)\sec \left(60°\right)-4\cos \left(60°\right)\cot \left(45°\right) \\ =3\times \frac{1}{2}\times 2-4\times \frac{1}{2}\times 1 \\ =3-2 \\ =1 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$