मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Cos 45° = ?

Advertisements
Advertisements

प्रश्न

cos 45° = ?

पर्याय

  • sin 45°

  • sec 45°

  • cot 45°

  • tan 45°

MCQ
Advertisements

उत्तर

sin 45°

Explanation:

`cos 45^circ = 1/sqrt2`, `sin 45^circ = 1/sqrt(2)`

∴ cos 45° = sin 45°

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Q.1 (A)

संबंधित प्रश्‍न

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`sqrt((1+sinA)/(1-sinA)) = secA + tanA`


Prove that `cosA/(1+sinA) + tan A =  secA`


Prove that `sqrt(sec^2 theta + cosec^2 theta) = tan theta + cot theta`


Prove the following trigonometric identities.

sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1


Prove the following trigonometric identities.

`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`


Prove the following identities:

`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`


Prove the following identities:

`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`


Prove the following identities:

`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`


If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`


Write the value of cos1° cos 2°........cos180° .


What is the value of (1 + cot2 θ) sin2 θ?


If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ? 


What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]


If \[\cos A = \frac{7}{25}\]  find the value of tan A + cot A. 


Prove the following identity : 

`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`


Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.


Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.


Prove the following identities.

`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ


Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)


If sin A = `1/2`, then the value of sec A is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×