English

Choose the correct alternative: sec2θ – tan2θ = ? - Geometry Mathematics 2

Advertisements
Advertisements

Question

Choose the correct alternative:

sec2θ – tan2θ =?

Options

  • 0

  • 1

  • 2

  • `sqrt(2)`

MCQ

Solution

1

1 + tan2θ = sec2θ

∵ sec2θ – tan2θ = 1.

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Trigonometry - Q.1 (A)

RELATED QUESTIONS

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


Prove the following trigonometric identities.

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


Prove the following trigonometric identities.

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


Prove the following trigonometric identities.

(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A


Prove that:

(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1


`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`


`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`


Write the value of `(sin^2 theta 1/(1+tan^2 theta))`. 


Prove the following identity : 

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


Prove the following identity : 

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


Prove the following identity :

`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`


Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.


If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2


Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`


Prove that sec2θ – cos2θ = tan2θ + sin2θ


Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1


If 2sin2β − cos2β = 2, then β is ______.


`sqrt((1 - cos^2theta) sec^2 theta) = tan theta` 


If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.


Find the value of sin2θ  + cos2θ

Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB2 + BC2 = `square`   .....(Pythagoras theorem)

Divide both sides by AC2

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×