Question

cot 9 + cot 81 + cot 27 + cot 63

Answer 1

cot 9° + cot 27° + cot 63° + cot 81°

= (cot9° + cot 81°) + (cot 27° + cot 63°)

= (sin81°cos9° + cos 81°sin9°)/(sin9°sin81°) + (sin63°cos27° + cos 63°sin27°)/(sin27°sin63°)

= sin90°/(sin9°sin81°) + sin90°//(sin27°sin63°

= 2/(cos72°) + 2/(cos36°) = 8/(√5 - 1) + 8/(√5 + 1) = 16√5/4 = 4√5.

Answer 2

$\cot 9+\cot 81+\cot 27+\cot 63=4\sqrt{5}$

Step-by-step explanation:

We have,

$\cot 9+\cot 81+\cot 27+\cot 63$

To find, the value of $\cot 9+\cot 81+\cot 27+\cot 63=?$

∴ $\cot 9+\cot 81+\cot 27+\cot 63$

$=(\cot 9+\cot 81)+(\cot 27+\cot 63)$

$=(\dfrac{\cos 9}{\sin 9} +\dfrac{\cos 81}{\sin 81})+(\dfrac{\cos 27}{\sin 27}+\dfrac{\cos 63}{\sin 63})$

Using trigonometric identity,

$\cot A=\dfrac{\cos A}{\sin A}$

$=\dfrac{\sin 81\cos 9+\cos 81\sin 9}{\sin 9\sin 81}+\dfrac{\sin 63\cos 27+\cos 63\sin 27}{\sin 27\sin 63}$

$=\dfrac{\sin (81+9)}{\sin 9\sin 81}+\dfrac{\sin (63+27)}{\sin 27\sin 63}$

Using trigonometric identity,

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

$=\dfrac{\sin 90}{\sin 9\sin 81}+\dfrac{\sin 90}{\sin 27\sin 63}$

$=\dfrac{1}{\sin 9\sin 81}+\dfrac{1}{\sin 27\sin 63}$

$=\dfrac{2}{\cos72}+\dfrac{2}{\cos36}$

$=\dfrac{2}{\dfrac{\sqrt{5}-1}{4}}+\dfrac{2}{\dfrac{\sqrt{5}-1}{4}}$

$=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}$

$=\dfrac{8(\sqrt{5}+1)+8(\sqrt{5}-1)}{4}$

$=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}$

Hence, $\cot 9+\cot 81+\cot 27+\cot 63=4\sqrt{5}$