हिंदी

(1 + cot^2A)/(1 + tan^2A) = ? A) tan^2θ B) sec^2θ C) cosec^2θ D) cot^2θ

Advertisements
Advertisements

प्रश्न

`(1 + cot^2A)/(1 + tan^2A)` = ?

विकल्प

  • tan2A

  • sec2A

  • cosec2A

  • cot2A

MCQ
Advertisements

उत्तर

cot2A

Explanation:

`(1 + cot^2A)/(1 + tan^2A)`

= `("cosec"^2A)/(sec^2A)`

= `(1/(sin^2A))/(1/(cos^2A))`

= `(cos^2A)/(sin^2A)`

= cot2A

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Trigonometry - Exercise

संबंधित प्रश्न

 Evaluate sin25° cos65° + cos25° sin65°


Evaluate without using trigonometric tables:

`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`


Prove the following trigonometric identities.

if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`


Prove the following identities:

`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`


Prove the following identities:

`cosA/(1 - sinA) = sec A + tan A`


Prove the following identities:

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


Prove that:

(cosec A – sin A) (sec A – cos A) sec2 A = tan A


`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`


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


If `cosec  theta = 2x and cot theta = 2/x ," find the value of"  2 ( x^2 - 1/ (x^2))`


\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to


If sin θ − cos θ = 0 then the value of sin4θ + cos4θ


Prove the following identity :

`(1 - cos^2θ)sec^2θ = tan^2θ`


Prove the following identity :

secA(1 + sinA)(secA - tanA) = 1


Prove the following identity :

`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`


Prove the following identity : 

`sqrt(cosec^2q - 1) = "cosq  cosecq"`


Prove the following identity :

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


Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.


Prove that `(1 + sec A)/(sec A) = (sin^2A)/(1 - cos A)`.


Prove the following identity:

(sin2θ – 1)(tan2θ + 1) + 1 = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×