Question

sin 18°/cos 72° + root 3 [tan10 tan 30 tan 40 tan 50 tan 80]
with out using the trigonometric table

Answer 1

sin18/cos (90 - 18 ) + root 3 ( tan (90 - 80)  tan (90 - 50)  tan 45  tan50 tan 80)
sin18/sin18 + root 3 ( cot 80 cot 50 tan 45 tan50 tan 80)
1 + root 3(1)
1 + 1 = 2 ANSWER.
note - it should be tan 45 instead of tan 40.
thanks...

Answer 2

Answer:

Concept:

Trigonometry

Step-by-step explanation:

Given:

$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left\{\tan 10^{\circ} \tan 40^{\circ} \tan 30^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right\}$

Find:

Solution for the above equation

Solution:

We know,

$\cos (90-\theta)=\sin \theta \\ \tan (90-\theta)=\cot \theta \\ \cot \theta=\frac{1}{\tan \theta} \\ \cos 72^{\circ}=\cos \left(90^{\circ}-18^{\circ}\right)=\sin 18^{\circ} \\ \tan 10^{\circ}=\tan \left(90^{\circ}-80^{\circ}\right)=\cot 80^{\circ} \\ \tan 40^{\circ}=\tan \left(90^{\circ}-50^{\circ}\right)=\cot 50^{\circ} \\ \csc 20^{\circ}=\sec \left(90^{\circ}-70^{\circ}\right)=\sec 70^{\circ}$

To solve,

$\begin{aligned}&\frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left(\tan 10^{\circ} \tan 40^{\circ} \tan 30^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right) \\&=\frac{\sin 18^{\circ}}{\sin 18^{\circ}}+\sqrt{3}\left(\cot 80^{\circ} \cot 50^{\circ} \tan 30^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right) \\&=1+\frac{\sqrt{3}\left(\tan 30^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right)}{\left(\tan 80^{\circ} \tan 50^{\circ}\right)} \\&=1+\sqrt{3}\left(\tan 30^{\circ}\right) \\\end{aligned}$

$\begin{aligned}\\&=1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right) \\\end{aligned}$

= 1 + 1

= 2

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