Evaluate :

Solution:

We have to evaluate the following values-

(i) We will use the values of known angles of different trigonometric ratios.

$=\frac{2}{3}\left(\cos ^{4} 30^{\circ}-\sin ^{4} 45^{\circ}\right)-3\left(\sin ^{2} 60^{\circ}-\sec ^{2} 45^{\circ}\right)+\frac{1}{4} \cot ^{2} 30^{\circ}$

$=\frac{2}{3}\left(\frac{9}{16}-\frac{1}{4}\right)-3\left(\frac{3}{4}-2\right)+\frac{1}{4}(3)$

$=\frac{2}{3}\left(\frac{5}{16}\right)+3\left(\frac{5}{4}\right)+\frac{3}{4}$

$=\frac{113}{24}$

(ii) We will use the values of known angles of different trigonometric ratios.

$=4\left(\sin ^{4} 30^{\circ}+\cos ^{4} 60^{\circ}\right)-\frac{2}{3}\left(\sin ^{2} 60^{\circ}-\cos ^{2} 45^{\circ}\right)+\frac{1}{2} \tan ^{2} 60^{\circ}$

$=4\left(\frac{1}{16}+\frac{1}{16}\right)-\frac{2}{3}\left(\frac{3}{4}-\frac{1}{2}\right)+\frac{1}{2}(3)$

$=4\left(\frac{1}{8}\right)-\frac{2}{3}\left(\frac{1}{4}\right)+\frac{3}{2}$

$=\frac{11}{6}$

(iii) We will use the properties of complementary angles.

$=\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\csc 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \csc 40^{\circ}$

$=\frac{\sin 50^{\circ}}{\sin 50^{\circ}}+\frac{\csc 40^{\circ}}{\csc 40^{\circ}}-4 \frac{\cos 50^{\circ}}{\cos 50^{\circ}}$

$=1+1-4$

$=-2$

(iv) We will use the properties of complementary angles.

$=\tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ}$

$=\cot 55^{\circ} \cot 50^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ}$

$=1$

(v) We will use the properties of complementary angles.

$=\csc \left(65^{\circ}+\theta\right)-\sec \left(25^{\circ}-\theta\right)-\tan \left(55^{\circ}-\theta\right)+\cot \left(35^{\circ}+\theta\right)$

$=\csc \left(65^{\circ}+\theta\right)-\csc \left(65^{\circ}+\theta\right)-\tan \left(55^{\circ}-\theta\right)+\tan \left(55^{\circ}-\theta\right)$

$=0$

(vi) We will use the properties of complementary angles.

$=\tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ}$

$=\cot 83^{\circ} \cot 67^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ}$

$=\sqrt{3}$

(vii) We will use the properties of complementary angles.

$=\frac{2 \sin 68^{\circ}}{\cos 22^{\circ}}-\frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}}-\frac{3 \tan 45^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}$

$=\frac{2 \sin 68^{\circ}}{\sin 68^{\circ}}-\frac{2 \cot 15^{\circ}}{5 \cot 15^{\circ}}-\frac{3 \tan 45^{\circ} \cot 70^{\circ} \cot 50^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}$

$=2-\frac{2}{5}-\frac{3}{5}$

$=1$

(viii) We will use the properties of complementary angles.

$=\frac{3 \cos 55^{\circ}}{7 \sin 35^{\circ}}-\frac{4\left(\cos 70^{\circ} \csc 20^{\circ}\right)}{7\left(\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}\right)}$

$=\frac{3 \cos 55^{\circ}}{7 \cos 55^{\circ}}-\frac{4\left(\cos 70^{\circ} \sec 70^{\circ}\right)}{7\left(\cot 85^{\circ} \cot 65^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}\right)}$

$=\frac{3}{7}-\frac{4}{7}$

$=-\frac{1}{7}$

(ix) We will use the properties of complementary angles.

$=\frac{\sin 18^{\circ}}{\sin 72^{\circ}}+\sqrt{3}(\tan 10 \tan 30 \tan 40 \tan 50 \tan 80)$

$=\frac{\sin 18^{\circ}}{\sin 18^{\circ}}+\sqrt{3}(\cot 80 \tan 30 \cot 50 \tan 50 \tan 80)$

$=1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)$

$=2$

(x) We will use the properties of complementary angles.

$=\frac{\cos 58^{\circ}}{\sin 32^{\circ}}+\frac{\sin 22^{\circ}}{\cos 68^{\circ}}-\frac{\cos 38^{\circ} \csc 52^{\circ}}{\tan 18^{0} \tan 35^{\circ} \tan 60^{\circ} \tan 72^{\circ} \tan 55^{\circ}}$

$=\frac{\cos 58^{\circ}}{\cos 58^{\circ}}+\frac{\sin 22^{\circ}}{\sin 22^{\circ}}-\frac{\cos 38^{\circ} \sec 38^{\circ}}{\cot 72^{\circ} \cot 55^{\circ} \tan 60 \tan 72^{\circ} \tan 55^{\circ}}$

$=1+1-\frac{1}{\sqrt{3}}$

$=\frac{6-\sqrt{3}}{3}$