Without trigonometric table- evaluate 63-tan 27-

The correct option is C 1 Taking 63^∘tan 27^∘ (90 27)^∘tan 27^∘ We know, #160; ( 90^0 θ #160;) #160; =tanθ #160; tan27^∘tan 27^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Complementary Trigonometric Ratios

Without trigo...

Question

Without trigonometric table, evaluate $\frac{cot 63^{\circ}}{tan 27^{\circ}}$

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1$
Taking
$\frac{cot 63^{\circ}}{tan 27^{\circ}}$
$\frac{cot (90 - 27)^{\circ}}{tan 27^{\circ}}$
We know,
$cot (90^{0} - θ) = tan θ$
$\frac{tan 27^{\circ}}{tan 27^{\circ}}$
$= 1$

Suggest Corrections

Similar questions

Q. Without using trigonometric tables, evaluate the following :

sec 41^{o} . sin 49^{o} + cos 49^{o} csc 41^{o} - \frac{2}{\sqrt{3}} (tan 20^{o} . tan 60^{o} . tan 70^{o})

Without using trigonometric tables, evaluate:
(i) $\frac{\sin 16 °}{\cos 74 °}$
(ii) $\frac{\sec 11 °}{cosec 79 °}$
(iii) $\frac{\tan 27 °}{\cot 63 °}$
(iv) $\frac{\cos 35 °}{\sin 55 °}$
(v) $\frac{cosec 42 °}{\sec 48 °}$
(vi) $\frac{\cot 38 °}{\tan 52 °}$

Q. Without using trigonometric tables, evaluate the follow

{s i n}^{2} 35^{0} + {s i n}^{2} 55^{0}

Q. Without using trigonometric tables, evaluate

{cosec}^{2} 57^{0} - {tan}^{2} 33^{0} + cos 44^{0} cosec 46^{0} - \sqrt{2} cos 45^{0} - {tan}^{2} 60^{0}

Without using trigonometric table, evaluate :

\frac{2}{3} c o s e c^{2} 58^{o} - \frac{2}{3} c o t 58^{o} t a n 32^{o} - \frac{5}{3} t a n 13^{o} t a n 37^{o} t a n 45^{o} t a n 53^{o} t a n 77^{o}

Join BYJU'S Learning Program

Without trigonometric table, evaluate cot63∘tan27∘

The correct option is C 1Takingcot63∘tan27∘cot(90−27)∘tan27∘We know, cot(900−θ)=tanθtan27∘tan27∘=1

Without trigonometric table, evaluate $\frac{cot 63^{\circ}}{tan 27^{\circ}}$

The correct option is C $1$
Taking
$\frac{cot 63^{\circ}}{tan 27^{\circ}}$
$\frac{cot (90 - 27)^{\circ}}{tan 27^{\circ}}$
We know,
$cot (90^{0} - θ) = tan θ$
$\frac{tan 27^{\circ}}{tan 27^{\circ}}$
$= 1$