# Without using trigonometric tables, prove that:

Question:

Without using trigonometric tables, prove that:
(i) sin212° + sin2 78° = 1
(ii) sec229° – cot261° = 1
(iii) tan256° – cot234° = 0
(iv) cos257° – sin233° = 0
(v) sec250° – cot240° = 1
(vi) cosec272° – tan218° = 1

Solution:

(i) To prove: sin212° + sin2 78° = 1

$\sin ^{2} 12^{\circ}+\sin ^{2} 78^{\circ}$

$=\left(\sin \left(90^{\circ}-78^{\circ}\right)\right)^{2}+\sin ^{2} 78^{\circ}$

$=\cos ^{2} 78^{\circ}+\sin ^{2} 78^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right)$

$=1 \quad$ (using the identity : $\cos ^{2} \theta+\sin ^{2} \theta=1$ )

Hence, $\sin ^{2} 12^{\circ}+\sin ^{2} 78^{\circ}=1$.

(ii) To prove: sec229° – cot261° = 1

$\sec ^{2} 29^{\circ}-\cot ^{2} 61^{\circ}$

$=\sec ^{2} 29^{\circ}-\left(\cot \left(90^{\circ}-29^{\circ}\right)\right)^{2}$

$=\sec ^{2} 29^{\circ}-\tan ^{2} 29^{\circ} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right)$

$=1 \quad$ (using the identity : $\sec ^{2} \theta-\tan ^{2} \theta=1$ )

Hence, $\sec ^{2} 29^{\circ}-\cot ^{2} 61^{\circ}=1$

(iii) To prove: tan256° – cot234° = 0

$\tan ^{2} 56^{\circ}-\cot ^{2} 34^{\circ}$

$=\left(\tan \left(90^{\circ}-34^{\circ}\right)\right)^{2}-\cot ^{2} 34^{\circ}$

$=\cot ^{2} 34^{\circ}-\cot ^{2} 34^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=0$

Hence, $\tan ^{2} 56^{\circ}-\cot ^{2} 34^{\circ}=0$.

(iv) To prove: cos257° – sin233° = 0

$\cos ^{2} 57^{\circ}-\sin ^{2} 33^{\circ}$

$=\left(\cos \left(90^{\circ}-33^{\circ}\right)\right)^{2}-\sin ^{2} 33^{\circ}$

$=\sin ^{2} 33^{\circ}-\sin ^{2} 33^{\circ} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=0$

Hence, $\cos ^{2} 57^{\circ}-\sin ^{2} 33^{\circ}=0$.

(v) To prove: sec250° – cot240° = 1

$\sec ^{2} 50^{\circ}-\cot ^{2} 40^{\circ}$

$=\sec ^{2} 50^{\circ}-\left(\cot \left(90^{\circ}-50^{\circ}\right)\right)^{2}$

$=\sec ^{2} 50^{\circ}-\tan ^{2} 50^{\circ} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right)$

$=1 \quad$ (using the identity : $\sec ^{2} \theta-\tan ^{2} \theta=1$ )

Hence, $\sec ^{2} 50^{\circ}-\cot ^{2} 40^{\circ}=1$.

(vi) To prove: cosec272° – tan218° = 1

$\operatorname{cosec}^{2} 72^{\circ}-\tan ^{2} 18^{\circ}$

$=\left(\operatorname{cosec}\left(90^{\circ}-18^{\circ}\right)\right)^{2}-\tan ^{2} 18^{\circ}$

$=\sec ^{2} 18^{\circ}-\tan ^{2} 18^{\circ} \quad\left(\because \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

$=1 \quad$ (using the identity : $\sec ^{2} \theta-\tan ^{2} \theta=1$ )

Hence, $\operatorname{cosec}^{2} 72^{\circ}-\tan ^{2} 18^{\circ}=1$