Without using trigonometric tables, prove that:
Question:

Without using trigonometric tables, prove that:
(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2

(iv) $\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}=\sqrt{3}$

(v) tan48° tan23° tan42° tan67° tan 45° = 1
(vi) (sin72° + cos18°)(sin72° − cos18°) = 0
(vii) cosec 39° cos 51° + tan 21° cot 69° – sec221° = 0

 

Solution:

(i) LHS $=\sin 53^{0} \cos 37^{0}+\cos 53^{0} \sin 37^{0}$

$=\sin \left(90^{\circ}-37^{0}\right) \cos 37^{0}+\cos \left(90^{0}-37^{0}\right) \sin 37^{0}$

$=\cos 37^{0} \cos 37^{0}+\sin 37^{0} \sin 37^{0}$

$=\cos ^{2} 37^{0}+\sin ^{2} 37^{0}$

$=1$

$=\mathrm{RHS}$

(ii) $\mathrm{LHS}=\cos 54^{0} \cos 36^{\circ}-\sin 54^{0} \sin 36^{0}$

$=\cos \left(90^{\circ}-36^{0}\right) \cos 36^{0}-\sin \left(90^{\circ}-36^{0}\right) \sin 36^{0}$

$=\sin 36^{0} \cos 36^{0}-\cos 36^{0} \sin 36^{0}$

$=0$

$=\mathrm{RHS}$

(iii) LHS $=\sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}$

$=\sec \left(90^{\circ}-20^{\circ}\right) \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right)$

$=\operatorname{cosec} 20^{\circ} \cdot \frac{1}{\operatorname{cosec} 20^{\circ}}+\frac{1}{\sec 20^{\circ}} \cdot \sec 20^{\circ}$

$=1+1$

$=2$

$=\mathrm{RHS}$

(iv) To prove: $\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}=\sqrt{3}$

$\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}$

$=\tan 15^{\circ} \tan 60^{\circ} \tan \left(90^{\circ}-15^{\circ}\right)$

$=\tan 15^{\circ} \tan 60^{\circ} \cot 15^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=\tan 15^{\circ} \tan 60^{\circ} \frac{1}{\tan 15^{\circ}} \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$

$=\tan 60^{\circ}$

$=\sqrt{3} \quad\left(\because \tan 60^{\circ}=\sqrt{3}\right)$

Hence, $\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}=\sqrt{3}$.

(v) LHS $=\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

$=\cot \left(90^{\circ}-48^{\circ}\right) \cot \left(90^{\circ}-23^{\circ}\right) \tan 42^{\circ} \tan 67^{\circ}$

$=\cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

$=\frac{1}{\tan 42^{\circ}} \times \frac{1}{\tan 67^{\circ}} \times \tan 42^{\circ} \times \tan 67^{\circ}$

$=1$

$=$ RHS

(vi) LHS $=\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)$

$=\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left[\cos \left(90^{\circ}-72^{\circ}\right)-\cos 18^{\circ}\right]$

$=\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\cos 18^{\circ}-\cos 18^{\circ}\right)$

$=\left(\sin 72^{\circ}+\cos 18^{\circ}\right)(0)$

$=0$

$=$ RHS

(vii) To prove: cosec39° cos51° + tan21° cot69° – sec221° = 0

$\operatorname{cosec} 39^{\circ} \cos 51^{\circ}+\tan 21^{\circ} \cot 69^{\circ}-\sec ^{2} 21^{\circ}$

$=\operatorname{cosec}\left(90^{\circ}-51^{\circ}\right) \cos 51^{\circ}+\tan 21^{\circ} \cot \left(90^{\circ}-21^{\circ}\right)-\sec ^{2} 21^{\circ}$

$=\sec 51^{\circ} \cos 51^{\circ}+\tan 21^{\circ} \tan 21^{\circ}-\sec ^{2} 21^{\circ} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right.$ and $\left.\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

$=\sec 51^{\circ} \cos 51^{\circ}+\tan ^{2} 21^{\circ}-\sec ^{2} 21^{\circ}$

$=\sec 51^{\circ} \cos 51^{\circ}-\left(\sec ^{2} 21^{\circ}-\tan ^{2} 21^{\circ}\right) \quad$ (using the identity : $\sec ^{2} \theta-\tan ^{2} \theta=1$ )

$=\sec 51^{\circ} \cos 51^{\circ}-1$

$=\frac{1}{\cos 51^{\circ}} \cos 51^{\circ}-1 \quad\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$

$=1-1$

$=0$

Hence, $\operatorname{cosec} 39^{\circ} \cos 51^{\circ}+\tan 21^{\circ} \cot 69^{\circ}-\sec ^{2} 21^{\circ}=0$

 

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