The correct options are A 1 B cos 72^∘/sin 18^∘ sin 18^∘/cos 72^∘ = sin (90^∘ 72^∘)/cos 72^∘ As, sin (90^∘ A) = cos A, the expression becomes cos 72^∘/cos ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

sin 18∘/cos 2...

Question

$\frac{sin 18^{\circ}}{cos 72^{\circ}} =$ ___

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$\frac{cos 72^{\circ}}{sin 18^{\circ}}$

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$tan 72^{\circ}$

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Solution

The correct options are
A 1
B
$\frac{cos 72^{\circ}}{sin 18^{\circ}}$

$\frac{sin 18^{\circ}}{cos 72^{\circ}}$
$= \frac{sin (90^{\circ} - 72^{\circ})}{cos 72^{\circ}}$
As, $sin (90^{\circ} - A) = cos A$ ,
the expression becomes
$\frac{cos 72^{\circ}}{cos 72^{\circ}} = 1$

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Similar questions

Prove that:

(i) $\frac{s i n 70^{\circ}}{c o s 20^{\circ}} + \frac{c o s e c 20^{\circ}}{s e c 70^{\circ}} - 2 c o s 70^{\circ} c o s e c 20^{\circ} = 0$

(ii) $\frac{c o s 80^{\circ}}{s i n 10^{\circ}} + c o s 59^{\circ} c o s e c 31^{\circ} = 2$

(iii) $\frac{2 s i n 68^{\circ}}{c o s 22^{\circ}} - \frac{2 c o t 15^{\circ}}{5 t a n 75^{\circ}} - \frac{3 t a n 45^{\circ} t a n 20^{\circ} t a n 40^{\circ} t a n 50^{\circ} t a n 70^{\circ}}{5} = 1$

(iv) $\frac{s i n 18^{\circ}}{c o s 72^{\circ}} + \sqrt{3} (t a n 10^{\circ} t a n 30^{\circ} t a n 40^{\circ} t a n 50^{\circ} t a n 80^{\circ}) = 2$

(v) $\frac{7 c o s 55^{\circ}}{3 s i n 35^{\circ}} - \frac{4 (c o s 70^{\circ} c o s e c 20^{\circ})}{3 (t a n 5^{\circ} t a n 25^{\circ} t a n 45^{\circ} t a n 65^{\circ} t a n 85^{\circ})} = 1$

$\frac{s i n 18^{\circ}}{c o s 72^{\circ}}$ ___

$\frac{c o s 22^{\circ} - s i n 22^{\circ}}{c o s 22^{\circ} + s i n 22^{\circ}}$ equal to tanA, $0^{\circ}$ < A < $90^{\circ}$ . Find the value of A.

___

\frac{s i n 18^{\circ}}{c o s 72^{\circ}} =

Q. Evaluate

\frac{sin 18^{\circ}}{cos 72^{\circ}}

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sin18∘cos72∘= ___

The correct options are A 1 B cos72∘sin18∘ sin18∘cos72∘ =sin(90∘−72∘)cos72∘ As, sin(90∘−A)=cosA, the expression becomes cos72∘cos72∘=1

$\frac{sin 18^{\circ}}{cos 72^{\circ}} =$ ___

The correct options are
A 1
B
$\frac{cos 72^{\circ}}{sin 18^{\circ}}$

$\frac{sin 18^{\circ}}{cos 72^{\circ}}$
$= \frac{sin (90^{\circ} - 72^{\circ})}{cos 72^{\circ}}$
As, $sin (90^{\circ} - A) = cos A$ ,
the expression becomes
$\frac{cos 72^{\circ}}{cos 72^{\circ}} = 1$