At - 90- which of the two angles are not defined- tan- cosec-cot- sin -cot- tan-tan- sec-

The correct option is D tan θ, sec θ As tan θ = sin θ/cos θ and at θ = 90^∘, cos 90^∘ = 0. So, tan θ is not defined. Similarly, sec θ =1/cos θ. So, sec θ is ...

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Byju's Answer

Standard IX

Mathematics

Trigonometric Ratio(sec,cosec,cot)

At θ = 90^∘, ...

Question

At $θ = 90^{\circ}$ , which of the two angles are not defined?

$c o t θ, s i n θ$

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$t a n θ, c o s e c θ$

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$c o t θ, t a n θ$

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$t a n θ, s e c θ$

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Solution

The correct option is D
$t a n θ, s e c θ$

As $t a n θ = \frac{s i n θ}{c o s θ}$ and at $θ = 90^{\circ}, c o s 90^{\circ} = 0$ . So, tan $θ$ is not defined.

Similarly, $s e c θ = \frac{1}{c o s θ}$ . So, $s e c θ$ is not defined at $θ = 90^{\circ}$ .

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Similar questions

Q. Prove that
(i)

\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1

(ii)

\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2

(iii)

\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1

(iv)

\"{\"1 + cotθ - \sec (\frac{π}{2} + θ)\"}\" \"{\"1 + cotθ + \sec (\frac{π}{2} + θ)\"}\" = 2 cotθ

(v)

\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1

Q. Find the value of

\frac{\cos (90 ° - θ) \cos θ}{\cot θ} - 1

Q. (i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

Q. (sinθ-cosecθ) (cosθ-secθ) (tanθ+cotθ) = 1

Q. Prove: (cosecΘ - sinΘ)(secΘ - cosΘ) = 1/tanΘ + cotΘ

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At θ=90∘, which of the two angles are not defined?

The correct option is D tan θ,sec θ As tan θ=sin θcos θ and at θ=90∘,cos 90∘=0. So, tan θ is not defined. Similarly, sec θ=1cos θ. So, sec θ is not defined at θ=90∘.

At $θ = 90^{\circ}$ , which of the two angles are not defined?

The correct option is D
$t a n θ, s e c θ$

As $t a n θ = \frac{s i n θ}{c o s θ}$ and at $θ = 90^{\circ}, c o s 90^{\circ} = 0$ . So, tan $θ$ is not defined.

Similarly, $s e c θ = \frac{1}{c o s θ}$ . So, $s e c θ$ is not defined at $θ = 90^{\circ}$ .