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प्रश्न
Which is not correct formula?
पर्याय
1 + tan2θ = sec2θ
1 + sec2θ = tan2θ
cosec2θ – cot2θ = 1
sin2θ + cos2θ = 1
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उत्तर
1 + sec2θ = tan2θ
Explanation:
(A) 1 + tan2θ = sec2θ: Correct. This is a fundamental Pythagorean identity.
(B) 1 + sec2θ = tan2θ: Incorrect. Rearranging the correct identity from (A) gives sec2θ – 1 = tan2θ.
(C) cosec2θ – cot2θ = 1: Correct. This is derived from the standard identity 1 + cot2θ = cosec2θ.
(D) sin2θ + cos2θ = 1: Correct. This is the primary Pythagorean trigonometric identity.
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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.
tan2 θ − sin2 θ = tan2 θ sin2 θ
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ
If sin A = `1/2`, then the value of sec A is ______.
