Advertisements
Advertisements
प्रश्न
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
Advertisements
उत्तर
L.H.S:
`tanA/(1 + sec A) - tanA/(1 - sec A)`
Taking LCM of the denominators,
= `(tanA(1 - sec A) - tanA(1 + sec A))/((1 + sec A)(1 - sec A))`
Since, (1 + sec A)(1 – sec A) = 1 – sec2A
= `(tan A(1 - secA - 1 - sec A))/(1 - sec^2A)`
= `(tan A(-2 sec A))/(1 - sec^2 A)`
= `(2 tan A *sec A)/(sec^2 A - 1)`
Since,
sec2A – tan2A = 1
sec2A – 1 = tan2A
= `(2 tan A * sec A)/(tan^2 A)`
Since, sec A = `(1/cosA)` and tan A = `(sinA/cosA)`
= `(2secA)/tanA = (2cosA)/(cosA sinA)`
= `2/sinA`
= 2 cosec A ...`(∵ 1/sinA = "cosec" A)`
= R.H.S
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
Choose the correct alternative:
sec2θ – tan2θ =?
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
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`
