हिंदी

`Cos^2 Theta + 1/((1+ Cot^2 Theta )) =1` - Mathematics

Advertisements
Advertisements

प्रश्न

`cos^2 theta + 1/((1+ cot^2 theta )) =1`

     

Advertisements

उत्तर

LHS = `cos^2 theta + 1/((1+cot^2 theta))`

      =` cos^2 theta + 1/(cosec^2  theta)`

      =` cos^2  theta + sin^2 theta`

      =1

      =RHS

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Trigonometric Identities - Exercises 1

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 8 Trigonometric Identities
Exercises 1 | Q 5.3

संबंधित प्रश्न

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z c tan θ, show that `x^2/a^2 + y^2/b^2 - x^2/c^2 = 1`


Prove the following identities:

`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`


Prove the following identities:

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


Show that : tan 10° tan 15° tan 75° tan 80° = 1


Prove that:

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


`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`


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


If cos A + cos2 A = 1, then sin2 A + sin4 A =


Prove the following identity:

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


Prove the following identity :

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


Prove the following identity : 

`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`


If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`


Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`


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


Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.


If tan α = n tan β, sin α = m sin β, prove that cos2 α  = `(m^2 - 1)/(n^2 - 1)`.


If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to


If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1


Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.


Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×