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प्रश्न
`(1-cos^2theta) sec^2 theta = tan^2 theta`
उत्तर
LHS = `(1-cos^2 theta)sec^2 theta`
=`sin^2 theta xx sec^2 theta (∵ sin^2 theta + cos^2 theta = 1)`
= `sin^2 theta xx 1/(cos^2 theta)`
=`(sin^2 theta)/(cos^2 theta)`
=`tan^2 theta`
=RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
`sin theta/((cot theta + cosec theta)) - sin theta /( (cot theta - cosec theta)) =2`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
If sinθ = `11/61`, find the values of cosθ using trigonometric identity.
What is the value of (1 + cot2 θ) sin2 θ?
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
Prove the following identity :
sinθcotθ + sinθcosecθ = 1 + cosθ
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
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 sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
