Fill in the blanks.

Solution:

(i) cot34° – tan56° = .......

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

$=\cot \left(90^{\circ}-56^{\circ}\right)-\tan 56^{\circ}$

$=\tan 56^{\circ}-\tan 56^{\circ} \quad\left(\because \cot \left(90^{\circ}-\theta\right)=\tan \theta\right)$

$=0$

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

(ii) cosec31° – sec59° = .......

$\cos ^{2} 67^{\circ}+\cos ^{2} 23^{\circ}$

$=\left(\cos \left(90^{\circ}-23^{\circ}\right)\right)^{2}+\cos ^{2} 23^{\circ} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=\sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ} \quad\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=1$

Hence, $\cos ^{2} 67^{\circ}+\cos ^{2} 23^{\circ}=\underline{1}$.

(iv) cosec254° – tan236° = .........

$\operatorname{cosec}^{2} 54^{\circ}-\tan ^{2} 36^{\circ}$

$=\left(\operatorname{cosec}\left(90^{\circ}-36^{\circ}\right)\right)^{2}-\tan ^{2} 36^{\circ} \quad\left(\because \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

$=\sec ^{2} 36^{\circ}-\tan ^{2} 36^{\circ} \quad\left(\because \sec ^{2} \theta-\tan ^{2} \theta=1\right)$

$=1$

Hence, $\operatorname{cosec}^{2} 54^{\circ}-\tan ^{2} 36^{\circ}=\underline{1} .$

(v) sec240° – cot250° = ..........

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

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

$=\operatorname{cosec}^{2} 50^{\circ}-\cot ^{2} 50^{\circ} \quad\left(\because \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right)$

$=1$

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