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प्रश्न
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
विकल्प
cos2 θ – sin2 θ = 1
cosec2 θ – sec2 θ = 1
sec2 θ – tan2 θ = 1
cot2 θ – tan2 θ = 1
उत्तर
sec2 θ – tan2 θ = 1
Explanation:
∵ sec2 θ = 1 + tan2 θ
∴ sec2 θ – tan2 θ = 1
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identities.
`tan theta + 1/tan theta = sec theta cosec theta`
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
`sin^2 theta + 1/((1+tan^2 theta))=1`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Prove that `(cot A - cos A)/(cot A + cos A) = (cos^2 A)/(1 + sin A)^2`
