Advertisements
Advertisements
Question
Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.
Advertisements
Solution
L.H.S. = `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1)`
= `(sin θ)/(1/cos θ + 1) + (sin θ)/(1/(cos θ) - 1`
= `(sin θ)/((1 + cos θ)/(cos θ)) + (sin θ)/((1 - cos θ)/(cos θ))`
= `(sin θ cos θ)/(1 + cos θ) + (sin θ cos θ)/(1 - cos θ)`
= `sin θ cos θ (1 /(1 + cos θ) + 1/(1 - cos θ))`
= `sin θ cos θ [(1 - cos θ + 1 + cos θ)/((1 + cos θ)(1 - cos θ))]`
= `sin θ cos θ (2/(1 - cos^2θ))` ...[∵ (a + b)(a – b) = a2 – b2]
= `sin θ cos θ xx 2/(sin^2θ)` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
= `2 xx (cos θ)/(sin θ)`
= 2 cot θ
= R.H.S.
∴ `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
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)
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
If cosA + cos2A = 1, then sin2A + sin4A = 1.
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
(1 – cos2 A) is equal to ______.
