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

Solution:

(i) We have $\sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$. So

$\sin 59^{\circ}+\cos 56^{\circ}=\sin \left(90^{\circ}-31^{\circ}\right)+\cos 90^{\circ}\left(90^{\circ}-34^{\circ}\right)$

$=\cos 31^{\prime \prime}+\sin 34$

Thus the desired expression is $\cos 31^{\circ}+\sin 34^{\circ}$

(ii) We know $\tan \left(90^{\circ}-\theta\right)=\cot \theta$ and $\cot \left(90^{\circ}-\theta\right)=\tan \theta$.So

$\tan 65^{\circ}+\cot 49^{\circ}=\tan \left(90^{\circ}-25^{\circ}\right)+\cot 90^{\circ}\left(90^{\circ}-41^{\circ}\right)$

$=\cot 25^{\circ}+\tan 41^{\circ}$

Thus the desired expression is $\cot 25^{\circ}+\tan 41^{\circ}$

(iii) We know that $\sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta$ and $\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta$. So

$\sec 76^{\circ}+\operatorname{cosec} 52^{\circ}=\sec \left(90^{\circ}-14^{\circ}\right)+\operatorname{cosec}\left(90^{\circ}-38^{\circ}\right)$

$=\operatorname{cosec} 14^{\prime \prime}+\sec 38^{\prime \prime}$

Thus the desired expression is $\operatorname{cosec} 14^{\circ}+\sec 38^{\circ}$

(iv) We know $\sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$

$\cos 78^{\circ}+\sec 78^{\circ}=\cos \left(90^{\circ}-12^{\circ}\right)+\sec \left(90^{\circ}-12^{\circ}\right)$

$=\sin 12^{\circ}+\operatorname{cosec} 12^{\circ}$

Thus the desired expression is $\sin 12^{\circ}+\operatorname{cosec} 12^{\circ}$

(v) We know $\sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta$. So

$\operatorname{cosec} 54^{\circ}+\sin 72^{\circ}=\operatorname{cosec}\left(90^{\circ}-36^{\circ}\right)+\sin \left(90^{\circ}-18^{\circ}\right)$

Thus the desired expression is $\sec 36^{\circ}+\cos 18^{\circ}$

(vi) We know that $\cot \left(90^{\circ}-\theta\right)=\tan \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta .$ So

$\cot 85^{\circ}+\cos 75^{\circ}=\cot \left(90^{\circ}-5^{\circ}\right)+\cos \left(90^{\circ}-15^{\circ}\right)$

$=\tan 5^{\circ}+\sin 15^{\circ}$

Thus the desired expression is $\tan 5^{\circ}+\sin 15^{\circ}$

(vii) We know that $\sin \left(90^{\circ}-\theta\right)=\cos \theta$ and $\cos \left(90^{\circ}-\theta\right)=\sin \theta$. So

$\sin 67^{\circ}+\cos 75^{\circ}=\sin \left(90^{\circ}-23^{\circ}\right)+\cos \left(90^{\circ}-15^{\circ}\right)$

$=\cos 23^{\circ}+\sin 15^{\circ}$

Thus the desired expression is $\cos 23^{\circ}+\sin 15^{\circ}$

$=\sec 36^{\circ}+\cos 18^{\circ}$