At - 90- which of the following trigonometric ratios are not defined-cot- sin -cot- tan-tan- cosec-tan- sec-

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

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Standard X

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At θ = 90^∘ w...

Question

At $θ = 90^{\circ}$ which of the following trigonometric ratios 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 at $θ = 90^{\circ}$ .

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

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Similar questions

Q. Prove:

(tan + cot θ + sec θ) (tan + cot θ - sec θ) = c o {sec}^{2} θ

Q. If

\sec θ = \frac{5}{4}

, find the value of

\frac{\sin θ - 2 \cos θ}{\tan θ - \cot θ}

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. How to prove this identity.:

17 . \frac{(1 + \tan^{2} θ) \sin^{2} θ}{\tan θ} = \tan θ

In the table, trigonometric ratios and the intervals of $θ$ are given. Match the intervals with the ratios which are positive in those intervals.

\begin{matrix} θ & gives positive values p. (0, \frac{π}{2}) & 1. Only sin θ, cosec θ q. (\frac{π}{2}, π) & 2. Only cos θ, sec θ r. (π, \frac{3 π}{2}) & 3. Only tan θ, cot θ s. (\frac{3 π}{2}, 2 π) & 4. All sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ \end{matrix}

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At θ=90∘ which of the following trigonometric ratios are not defined?

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

At $θ = 90^{\circ}$ which of the following trigonometric ratios 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 at $θ = 90^{\circ}$ .

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