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Prove that:i ...

Question

Prove that:
(i) tan 720° − cos 270° − sin 150° cos 120° = $\frac{1}{4}$
(ii) sin 780° sin 480° + cos 120° sin 150° = $\frac{1}{2}$
(iii) sin 780° sin 120° + cos 240° sin 390 = $\frac{1}{2}$
(iv) sin 600° cos 390° + cos 480° sin 150° = −1
(v) tan 225° cot 405° + tan 765° cot 675° = 0

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Solution

$(i) LHS = \tan (720 °) - \cos (270 °) - \sin (150 °) \cos (120 °) = \tan (90 ° \times 8 + 0 °) - \cos (90 ° \times 3 + 0 °) - \sin (90 ° \times 1 + 60 °) \cos (90 ° \times 1 + 30 °) = \tan (0 °) - \sin (0 °) - \cos (60 °) [- \sin (30 °)] = \tan (0 °) - \sin (0 °) + \cos (60 °) \sin (30 °) = 0 - 0 + \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = RHS Hence proved .$

$(ii) LHS = \sin (780 °) \sin (480 °) + \cos (120 °) \sin (150 °) = \sin (90 ° \times 8 + 60 °) \sin (90 ° \times 5 + 30 °) + \cos (90 ° \times 1 + 30 °) \sin (90 ° \times 1 + 60 °) = \sin (60 °) \cos (30 °) + [- \sin (30 °)] \cos (60 °) = \sin (60 °) \cos (30 °) - \sin (30 °) \cos (60 °) = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = RHS Hence proved .$

$(iii) LHS = \sin (780 °) \sin (120 °) + \cos (240 °) \sin (390 °) = \sin (90 ° \times 8 + 60 °) \sin (90 ° \times 1 + 30 °) + \cos (90 ° \times 2 + 60 °) \sin (90 ° \times 4 + 30 °) = \sin 60 ° \cos 30 ° + [- \cos 60 °] \sin 30 ° = \sin 60 ° \cos 30 ° - \cos 60 ° \sin 30 ° = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = RHS Hence proved .$

$(iv) LHS = \sin 600 ° \cos 390 ° + \cos 480 ° \sin 150 ° = \sin (90 ° \times 6 + 60 °) \cos (90 ° \times 4 + 30 °) + \cos (90 ° \times 5 + 30 °) \sin (90 ° \times 1 + 60 °) = [- \sin 60 °] \cos 30 ° + [- \sin 30 °] \cos 60 ° = - \sin 60 ° \cos (30 °) - \sin 30 ° \cos 60 ° = - \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = - \frac{3}{4} - \frac{1}{4} = - 1 = RHS Hence proved .$

$(v) LHS = \tan 225 ° \cot 405 ° + \tan 765 ° \cot 675 ° = \tan (90 ° \times 2 + 45 °) \cot (90 ° \times 4 + 45 °) + \tan (90 ° \times 8 + 45 °) \cot (90 ° \times 7 + 45 °) = \tan 45 ° \cot 45 ° + \tan 45 ° [- \tan 45 °] = \tan 45 ° \cot 45 ° - \tan 45 ° \tan 45 ° = 1 \times 1 - 1 \times 1 = 1 - 1 = 0 = RHS Hence proved .$

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Similar questions

tan 720^{\circ} - cos 270^{\circ} - sin 150^{\circ} cos 120^{\circ} = \frac{1}{4}

sin 780^{\circ} sin 480^{\circ} + cos 120^{\circ} sin 150^{\circ} = \frac{1}{2}

which of the above statements are true.

Q. Prove that:
(i) tan 225° cot 405° + tan 765° cot 675° = 0
(ii)

\sin \frac{8 π}{3} \cos \frac{23 π}{6} + \cos \frac{13 π}{3} \sin \frac{35 π}{6} = \frac{1}{2}

(iii) cos 24° + cos 55° + cos 125° + cos 204° + cos 300° =

\frac{1}{2}

(iv) tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
(v) cos 570° sin 510° + sin (−330°) cos (−390°) = 0
(vi)

\tan \frac{11 π}{3} - 2 \sin \frac{4 π}{6} - \frac{3}{4} {cosec}^{2} \frac{π}{4} + 4 \cos^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}

3 \sin \frac{π}{6} \sec \frac{π}{3} - 4 \sin \frac{5 π}{6} \cot \frac{π}{4} = 1

Q. Prove that:
(i)

\tan 4 π - \cos \frac{3 π}{2} - \sin \frac{5 π}{6} \cos \frac{2 π}{3} = \frac{1}{4}

\sin \frac{13 π}{3} \sin \frac{8 π}{3} + \cos \frac{2 π}{3} \sin \frac{5 π}{6} = \frac{1}{2}

\sin \frac{13 π}{3} \sin \frac{2 π}{3} + \cos \frac{4 π}{3} \sin \frac{13 π}{6} = \frac{1}{2}

\sin \frac{10 π}{3} \cos \frac{13 π}{6} + \cos \frac{8 π}{3} \sin \frac{5 π}{6} = - 1

\tan \frac{5 π}{4} \cot \frac{9 π}{4} + \tan \frac{17 π}{4} \cot \frac{15 π}{4} = 0

cos 510^{\circ} cos 330^{\circ} + sin 390^{\circ} cos 120^{\circ} = - 1

Q. Find the value of

tan 225^{\circ} cot 405^{\circ} + tan 765^{\circ} cot 675^{\circ}

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