$\cot 12 ° \cot 38 ° \cot 52 ° \cot 60 ° \cot 78 °$

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Solution

Find the value of the given expression
$\cot 12 ° \cot 38 ° \cot 52 ° \cot 60 ° \cot 78 °$
$= \cot 12 ° \cot 38 ° \cot (90 ° - 38 °) \times \frac{1}{\sqrt{3}} \times \cot (90 ° - 12 °)$
$= \frac{1}{\tan 12 °} \times \frac{1}{\tan 38 °} \times \frac{1}{\sqrt{3}} \times \tan 38 ° \times \tan 12 °$
$= \frac{1}{\sqrt{3}}$

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{(\frac{\sin 49 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\sin 49 °})}^{2} = 2

Prove that:

(i) $t a n 5^{\circ} t a n 25^{\circ} t a n 30^{\circ} t a n 65^{\circ} t a n 85^{\circ} = \frac{1}{\sqrt{3}}$

(ii) $c o t 12^{\circ} c o t 38^{\circ} c o t 52^{\circ} c o t 60^{\circ} c o t 78^{\circ} = \frac{1}{\sqrt{3}}$

(iii) $c o s 15^{\circ} c o s 35^{\circ} c o s e c 55^{\circ} c o s 60^{\circ} c o s e c 75^{\circ} = \frac{1}{2}$

(iv) $c o s 1^{\circ} c o s 2^{\circ} c o s 3^{\circ} . . . c o s 180^{\circ} = 0$

(v) ${(\frac{s i n 49^{\circ}}{c o s 41^{\circ}})}^{2} + {(\frac{c o s 41^{\circ}}{s i n 49^{\circ}})}^{2} = 2$

c o t 12^{\circ} c o t 38^{\circ} c o t 52^{\circ} c o t 60^{\circ} c o t 78^{\circ} =

cos 12^{\circ} cot 38^{\circ} cot 52^{\circ} cot 60^{\circ} cot 78^{\circ}

cot 12 cot 38 cot 52 cot 60 cot 78

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