Without using trigonometric tables, prove that:

Solution:

(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, $\cos ^{2} 72^{\circ}+\cos ^{2} 18^{\circ}=1$.