English

Evaluate: (Tan 65°)/(Cot 25°) - Mathematics

Advertisements
Advertisements

Question

Evaluate:
`(tan 65°)/(cot 25°)`

Sum

Solution

`(tan 65°)/(cot 25°)`

= `(tan  90° - 25°)/(cot 25°)`     ...( tan(90°θ) = cotθ )

= `( cot 25° )/( cot 25°)`

= 1

shaalaa.com
  Is there an error in this question or solution?
2018-2019 (March) 30/4/3

RELATED QUESTIONS

Prove the following trigonometric identities.

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


Prove the following trigonometric identities

tan2 A + cot2 A = sec2 A cosec2 A − 2


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:

`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`


`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`


`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`


Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`

 


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


Write the value of tan1° tan 2°   ........ tan 89° .


Define an identity.


Write the value of sin A cos (90° − A) + cos A sin (90° − A).


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


Prove the following identity : 

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


Prove the following identity : 

`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`


Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`


If cos A = `(2sqrt("m"))/("m" + 1)`, then prove that cosec A = `("m" + 1)/("m" - 1)`


Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`


If tan α + cot α = 2, then tan20α + cot20α = ______.


Prove the following:

`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A


Prove the following that:

`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×