Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 60° − sec 30° = 0
(iv) cot 34° − tan 56° = 0
(v) sin248° + sin242° = 1
(vi) cos272° + cos218° = 1
(i) $\mathrm{LHS}=\cos 81^{0}-\sin 9^{0}$
$=\cos \left(90^{0}-9^{0}\right)-\sin 9^{0}$
$=\sin 9^{0}-\sin 9^{0}$
$=0$
$=\mathrm{RHS}$
(ii) LHS $=\tan 71^{\circ}-\cot 19^{\circ}$
$=\tan \left(90^{\circ}-19^{\circ}\right)-\cot 19^{0}$
$=\cot 19^{0}-\cot 19^{0}$
$=0$
(iii) To Prove: cosec60° − sec30° = 0
$\operatorname{cosec} 60^{\circ}-\sec 30^{\circ}$
$=\operatorname{cosec}\left(90^{\circ}-30^{\circ}\right)-\sec 30^{\circ}$
$=\sec 30^{\circ}-\sec 30^{\circ} \quad\left(\because \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$
$=0$
Hence, $\operatorname{cosec} 60^{\circ}-\sec 30^{\circ}=0$
(iv) To Prove: cot34° − tan56° = 0
$\cot 34^{\circ}-\tan 56^{\circ}$
$=\cot \left(90^{\circ}-56^{\circ}\right)-\tan 56^{\circ}$
$=\tan 56^{\circ}-\tan 56^{\circ} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right)$
$=0$
Hence, $\cot 34^{\circ}-\tan 56^{\circ}=0$.
(v) LHS $=\sin ^{2} 48^{0}+\sin ^{2} 42^{0}$
$=\sin ^{2}\left(90^{\circ}-42^{0}\right)+\sin ^{2} 42^{0}$
$=\cos ^{2} 42^{0}+\sin ^{2} 42^{0}$
$=1$
$=\mathrm{RHS}$
(vi) To Prove: cos272° + cos218° = 1
$\cos ^{2} 72^{\circ}+\cos ^{2} 18^{\circ}$
$=\left(\cos \left(90^{\circ}-18^{\circ}\right)\right)^{2}+\cos ^{2} 18^{\circ}$
$=\sin ^{2} 18^{\circ}+\cos ^{2} 18^{\circ} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$
$=1 \quad$ (using the identity : $\cos ^{2} \theta+\sin ^{2} \theta=1$ )
Hence, $\cos ^{2} 72^{\circ}+\cos ^{2} 18^{\circ}=1$.
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