Which of the following complementary trigonometric ratios are not defined at 0°?

C o t θ

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

C o s θ

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

C o s e c θ

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

S e c θ

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

Open in App

Solution

The correct options are
A $C o t θ$
C $C o s e c θ$
we know that,
$C o t 0 ° = N o t d e f i n e d C o s 0 ° = 1 C o s e c 0 ° = N o t d e f i n e d S e c 0 ° = 1$

$S o, C o t θ a n d C o s e c θ$ are the two complementary trigonometric ratios which are not defined at 0°.

Suggest Corrections

Similar questions

θ = 90^{\circ}

(i) \sin θ \cos θ - \frac{\sin θ \cos (90 ° - θ) \cos θ}{s e c (90 ° - θ)} - \frac{\cos θ \sin (90 ° - θ) \sin θ}{\cos e c (90 ° - θ)} = 0

2 {(\frac{\tan 35 °}{\cot 55 °})}^{2} + (\frac{\cot 55 °}{\tan 35 °}) - 3 (\frac{\sec 40 °}{cosec 50 °})

\frac{\sin 35 ° \cos 55 ° + \cos 35 ° \sin 55 °}{{cosec}^{2} 10 ° - \tan^{2} 80 °}

\frac{\cos θ}{\sin (90 ° - θ)} + \frac{\sin θ}{\cos (90 ° - θ)} = 2

θ \sin (90 ° - θ) + \sin θ \cos (90 ° - θ) = 1

\frac{\tan θ}{\tan (90 ° - θ)} + \frac{\sin (90 ° - θ)}{\cos θ} = \sec^{2} θ

\frac{\cos (90 ° - A) \sin (90 ° - A)}{\tan (90 ° - A)} = 1 - \cos^{2} A

\frac{\sin (90 ° - A)}{cosec (90 ° - A)} + \frac{\cos (90 ° - A)}{\sec (90 ° - A)} = 1

\frac{\cos θ}{\sin (90 ° - 0)} + \frac{\cos (90 ° - θ)}{\sec (90 ° - θ)} - 3 \tan^{2} 30 °

Join BYJU'S Learning Program