Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

(i) sin72° + cosec72°

$\sin 72^{\circ}+\operatorname{cosec} 72^{\circ}$

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

$=\cos 18^{\circ}+\sec 18^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

Hence, $\sin 72^{\circ}+\operatorname{cosec} 72^{\circ}=\cos 18^{\circ}+\sec 18^{\circ} .$

(ii) cosec66° + tan66°

$\operatorname{cosec} 66^{\circ}+\tan 66^{\circ}$

$=\operatorname{cosec}\left(90^{\circ}-24^{\circ}\right)+\tan \left(90^{\circ}-24^{\circ}\right)$

$=\sec 24^{\circ}+\cot 24^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right.$ and $\left.\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

Hence, $\operatorname{cosec} 66^{\circ}+\tan 66^{\circ}=\sec 24^{\circ}+\cot 24^{\circ} .$

(iii) tan68° + sec68°

$\tan 68^{\circ}+\sec 68^{\circ}$

$=\tan \left(90^{\circ}-22^{\circ}\right)+\sec \left(90^{\circ}-22^{\circ}\right)$

$=\cot 22^{\circ}+\operatorname{cosec} 22^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right.$ and $\left.\sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta\right)$

Hence, $\tan 68^{\circ}+\sec 68^{\circ}=\cot 22^{\circ}+\operatorname{cosec} 22^{\circ}$.

(iv) cot59° + cosec59°

$\cot 59^{\circ}+\operatorname{cosec} 59^{\circ}$

$=\cot \left(90^{\circ}-31^{\circ}\right)+\operatorname{cosec}\left(90^{\circ}-31^{\circ}\right)$

$=\tan 31^{\circ}+\sec 31^{\circ} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right.$ and $\left.\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

Hence, $\cot 59^{\circ}+\operatorname{cosec} 59^{\circ}=\tan 31^{\circ}+\sec 31^{\circ}$

(v) cos51° + cot49° – sec47°

$\cos 51^{\circ}+\cot 49^{\circ}-\sec 47^{\circ}$

$=\cos \left(90^{\circ}-39^{\circ}\right)+\cot \left(90^{\circ}-41^{\circ}\right)-\sec \left(90^{\circ}-43^{\circ}\right)$

$=\sin 39^{\circ}+\tan 41^{\circ}-\operatorname{cosec} 43^{\circ} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta, \cot \left(90^{\circ}-\theta\right)=\tan \theta\right.$ and $\left.\sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta\right)$

Hence, $\cos 51^{\circ}+\cot 49^{\circ}-\sec 47^{\circ}=\sin 39^{\circ}+\tan 41^{\circ}-\operatorname{cosec} 43^{\circ}$

(vi) sin67° + cos75°

$\sin 67^{\circ}+\cos 75^{\circ}$

$=\sin \left(90^{\circ}-23^{\circ}\right)+\cos \left(90^{\circ}-15^{\circ}\right)$

$=\cos 23^{\circ}+\sin 15^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

Hence, $\sin 67^{\circ}+\cos 75^{\circ}=\cos 23^{\circ}+\sin 15^{\circ}$.