हिंदी

In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.

Advertisements
Advertisements

प्रश्न

In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.

योग
Advertisements

उत्तर

sin θ = `("AB")/("AC")` and tan θ = `("AB")/("BC")`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2025-2026 (March) Model set 1 by shaalaa.com

संबंधित प्रश्न

In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine:

sin A, cos A


In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine:

sin C, cos C


 In Given Figure, find tan P – cot R.


State whether the following are true or false. Justify your answer.

The value of tan A is always less than 1.


State whether the following are true or false. Justify your answer.

cos A is the abbreviation used for the cosecant of angle A.


In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.

`sin A = 2/3`


In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

`tan alpha = 5/12`


In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.

`sec theta = 13/5`


if `tan theta = 12/13` Find `(2 sin theta cos theta)/(cos^2 theta - sin^2 theta)`


Evaluate the following

cos 60° cos 45° - sin 60° ∙ sin 45°


Evaluate the following

cos2 30° + cos2 45° + cos2 60° + cos2 90°


Evaluate the following

`2 sin^2 30^2 - 3 cos^2 45^2 + tan^2 60^@`


Evaluate the following:

(cosec2 45° sec2 30°)(sin2 30° + 4 cot2 45° − sec2 60°)


If sin (A − B) = sin A cos B − cos A sin B and cos (A − B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.


The value of cos 0°. cos 1°. cos 2°. cos 3°… cos 89° cos 90° is ______.


If A and (2A – 45°) are acute angles such that sin A = cos (2A – 45°), then tan A is equal to ______.


If 4 tanθ = 3, then `((4 sintheta - costheta)/(4sintheta + costheta))` is equal to ______.


Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.

Proof: L.H.S. = sec θ + tan θ

= `1/square + square/square`

= `square/square`  ......`(∵ sec θ = 1/square, tan θ = square/square)`

= `((1 + sin θ) square)/(cos θ  square)`  ......[Multiplying `square` with the numerator and denominator]

= `(1^2 - square)/(cos θ  square)`

= `square/(cos θ  square)`

= `cos θ/(1 - sin θ)` = R.H.S.

∴ L.H.S. = R.H.S.

∴ sec θ + tan θ = `cos θ/(1 - sin θ)`


If sin θ + cos θ = `sqrt(2)` then tan θ + cot θ = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×