हिंदी

If tan θ = 1312, then cot θ = ? - Geometry Mathematics 2

Advertisements
Advertisements

प्रश्न

If tan θ = `13/12`, then cot θ = ?

योग
Advertisements

उत्तर

 cot θ = `1/tantheta`

= `1/(13/12)`

∴ cot θ = `12/13`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Trigonometry - Q.1 (B)

संबंधित प्रश्न

(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.


If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.


Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`


Prove the following trigonometric identities.

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


Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Show that one of the values of each member of this equality is sin α sin β sin γ


Prove the following identities:

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


Prove the following identities:

`sinA/(1 - cosA) - cotA = cosecA`


If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`


Prove that:

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


If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1


If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`


What is the value of 9cot2 θ − 9cosec2 θ? 


If \[sec\theta + tan\theta = x\] then \[tan\theta =\] 


Prove the following identity :

 ( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ) 


Prove the following identity : 

`((1 + tan^2A)cotA)/(cosec^2A) = tanA`


Prove the following identity :

`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`


Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.


If x = h + a cos θ, y = k + b sin θ. 
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.


Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.


Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×