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Choose the correct alternative: cos θ. sec θ = ? - Geometry Mathematics 2

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प्रश्न

Choose the correct alternative:

cos θ. sec θ = ?

विकल्प

  • 1

  • 0

  • `1/2`

  • `sqrt(2)`

MCQ
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उत्तर

1

cos θ. sec θ = cos θ. `1/"cos θ"` = 1.

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

संबंधित प्रश्न

Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`


Prove the following trigonometric identities.

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


Prove the following trigonometric identities.

(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A


Prove the following identities:

`1/(tan A + cot A) = cos A sin A`


Prove the following identities:

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


If sec A + tan A = p, show that:

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


(i)` (1-cos^2 theta )cosec^2theta = 1`


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


`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`


Write the value of `(1 - cos^2 theta ) cosec^2 theta`.


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


If cosec θ − cot θ = α, write the value of cosec θ + cot α.


If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.


Prove the following identity :

`(1 - sin^2θ)sec^2θ = 1`


Without using trigonometric table , evaluate : 

`sin72^circ/cos18^circ  - sec32^circ/(cosec58^circ)`


Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0


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


Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ


If 2 cos θ + sin θ = `1(θ ≠ π/2)`, then 7 cos θ + 6 sin θ is equal to ______.


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` 


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