मराठी

Without using tables, evaluate the following: sec30° cosec60° + cos60° sin30°. - Mathematics

Advertisements
Advertisements

प्रश्न

Without using tables, evaluate the following: sec30° cosec60° + cos60° sin30°.

बेरीज

उत्तर

sec30° cosec60° + cos60° sin30°.

cos30° = `sqrt(3)/(2)`

⇒ sec30° = `(2)/sqrt(3)`

sin60° = `sqrt(3)/(2)`

⇒ cosec60° = `(2)/sqrt(3)`

cos60° = `(1)/(2) , sin30° = (1)/(2)`

sec30° cosec60° + cos60° sin30°

= `(2)/sqrt(3) xx (2)/sqrt(3) + (1)/(2) xx (1)/(2)`

= `(4)/(3) + (1)/(4)`

= `(16 + 3)/(12)`

= `(19)/(12)`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 27: Trigonometrical Ratios of Standard Angles - Exercise 27.1

APPEARS IN

फ्रँक Mathematics [English] Class 9 ICSE
पाठ 27 Trigonometrical Ratios of Standard Angles
Exercise 27.1 | Q 1.02

संबंधित प्रश्‍न

Find the value of θ in each of the following :

(i) 2 sin 2θ = √3      (ii) 2 cos 3θ = 1


Evaluate the following in the simplest form: sin 60º cos 45º + cos 60º sin 45º


Evaluate the following :

`cos 19^@/sin 71^@`


Evaluate the following :

`(cot 40^@)/cos 35^@ -  1/2 [(cos 35^@)/(sin 55^@)]`


Evaluate the following :

`(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)`


Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°

 sin 67° + cos 75°


Prove that sin 48° sec 42° + cos 48° cosec 42° = 2


Evaluate: `sin 18^@/cos 72^@  + sqrt3 [tan 10° tan 30° tan 40° tan 50° tan 80°]`


Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

sec78° + cosec56°


Find the value of:

tan2 30° + tan2 45° + tan2 60°


prove that:

sin (2 × 30°) = `(2 tan 30°)/(1+tan^2 30°)`


prove that:

cos (2 x 30°) = `(1 – tan^2 30°)/(1+tan^2 30°)`


If `sqrt3` = 1.732, find (correct to two decimal place)  the value of  `(2)/(tan 30°)`


Evaluate : 

`(3 sin 3"B" + 2 cos(2"B" + 5°))/(2 cos 3"B" – sin (2"B" – 10°)` ; when "B" = 20°.


Without using tables, evaluate the following: sin60° sin30°+ cos30° cos60°


Without using tables, evaluate the following: (sin90° + sin45° + sin30°)(sin90° - cos45° + cos60°).


Without using table, find the value of the following:

`(sin30° - sin90° +  2cos0°)/(tan30° tan60°)` 


Prove that: `((cot30° + 1)/(cot30° -1))^2 = (sec30° + 1)/(sec30° - 1)`


Find the value of x in the following: `sqrt(3)sin x` = cos x


Find the value of x in the following: `sqrt(3)`tan 2x = cos60° + sin45° cos45°


Without using table, find the value of the following: `(tan^2 60° + 4cos^2 45° + 3sec^2 30° + 5cos90°)/(cosec30° + sec60° - cot^2 30°)`


If A = 30° and B = 60°, verify that: cos (A + B) = cos A cos B - sin A sin B


If A = 30° and B = 60°, verify that: `(sin("A" -"B"))/(sin"A" . sin"B")` = cotB - cotA


If A = B = 45°, verify that sin (A - B) = sin A .cos B - cos A.sin B


If A = B = 45°, verify that cos (A − B) = cos A. cos B + sin A. sin B


If tan `"A" = (1)/(2), tan "B" = (1)/(3) and tan("A" + "B") = (tan"A" + tan"B")/(1 - tan"A" tan"B")`, find A + B.


Verify the following equalities:

cos 90° = 1 – 2sin2 45° = 2cos2 45° – 1


Verify cos3A = 4cos3A – 3cosA, when A = 30°


Evaluate: `(5  "cosec"^2  30^circ - cos 90^circ)/(4 tan^2 60^circ)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×