Advertisements
Advertisements
प्रश्न
Find the acute angle θ such that 5tan2θ + 3 = 9secθ.
Advertisements
उत्तर
∴ 5 tan2θ+ 3 = 9 secθ
∴ 5 (sec2θ – 1) + 3 = 9 secθ
∴ 5 sec2θ – 5 + 3 = 9 secθ
∴ 5 sec2θ – 9secθ − 2 = 0
∴ 5 sec2θ – 10secθ + secθ – 2 = 0
∴ 5 secθ (secθ – 2) + 1 (secθ – 2) = 0
∴ (secθ – 2)(5secθ + 1) = 0
∴ secθ – 2 = 0 or 5secθ + 1 = 0
∴ secθ = 2 or secθ = `-1/5`
But secθ ≥ 1 or secθ ≤ – 1 for all θ ∈ R, where cosθ ≠ 0
∴ secθ ≠ `-1/5`
∴ secθ = 2 = sec60° ...[∵ θ is acute angle]
∴ θ = 60°
APPEARS IN
संबंधित प्रश्न
Eliminate θ from the following :
x = 6cosecθ, y = 8cotθ
Eliminate θ from the following :
x = 5 + 6cosecθ, y = 3 + 8cotθ
Eliminate θ from the following:
2x = 3 − 4 tan θ, 3y = 5 + 3 sec θ
Find the acute angle θ such that 2 cos2θ = 3 sin θ.
Find sinθ such that 3cosθ + 4sinθ = 4
If cosecθ + cotθ = 5, then evaluate secθ.
Prove the following identities:
(cos2A – 1) (cot2A + 1) = −1
Prove the following identities:
(sinθ + sec θ)2 + (cosθ + cosec θ)2 = (1 + cosecθ sec θ)2
Prove the following identities:
`tan^3theta/(1 + tan^2theta) + cot^3theta/(1 + cot^2theta` = secθ cosecθ – 2sinθ cosθ
Prove the following identities:
`1/(sectheta + tantheta) - 1/costheta = 1/costheta - 1/(sectheta - tantheta)`
Prove the following identity:
`tantheta/(sectheta - 1) = (sectheta + 1)/tantheta`
Prove the following identities:
`cottheta/("cosec" theta - 1) = ("cosec" theta + 1)/cot theta`
Select the correct option from the given alternatives:
`tan"A"/(1 + sec"A") + (1 + sec"A")/tan"A"` is equal to
Select the correct option from the given alternatives:
If θ = 60°, then `(1 + tan^2theta)/(2tantheta)` is equal to
Select the correct option from the given alternatives:
If cosecθ + cotθ = `5/2`, then the value of tanθ is
Select the correct option from the given alternatives:
If cosecθ − cotθ = q, then the value of cot θ is
Select the correct option from the given alternatives:
The value of tan1°.tan2°tan3°..... tan89° is equal to
Prove the following:
`((1 + cot theta + tan theta)(sin theta - costheta)) /(sec^3theta - "cosec"^3theta)`= sin2θ cos2θ
Prove the following:
2 sec2θ – sec4θ – 2cosec2θ + cosec4θ = cot4θ – tan4θ
Prove the following:
sin4θ + cos4θ = 1 – 2 sin2θ cos2θ
Prove the following:
2(sin6θ + cos6θ) – 3(sin4θ + cos4θ) + 1 = 0
Prove the following:
sin4θ +2sin2θ . cos2θ = 1 − cos4θ
Prove the following:
(sinθ + cosecθ)2 + (cosθ + secθ)2 = tan2θ + cot2θ + 7
Prove the following:
(1 + tanA · tanB)2 + (tanA − tanB)2 = sec2A · sec2B
Prove the following:
`(1 + cottheta + "cosec" theta)/(1 - cottheta + "cosec" theta) = ("cosec" theta + cottheta - 1)/(cottheta - "cosec"theta + 1)`
Prove the following:
`("cosec"theta + cottheta + 1)/(cottheta + "cosec" theta - 1) = cottheta/("cosec"theta - 1)`
Prove the following:
`("cosec"theta + cottheta - 1)/( "cosec"theta + cot theta + 1) =(1-sintheta)/costheta`
Prove the following identity:
`(1 - sec theta + tan theta)/(1 + sec theta - tan theta) = (sec theta + tan theta - 1)/(sec theta + tan theta + 1)`
If θ lies in the first quadrant and 5 tan θ = 4, then `(5 sin θ - 3 cos θ)/(sin θ + 2 cos θ)` is equal to ______.
If 5 tan θ = 4. then `(5 sin θ − 3 cos θ)/(5 sin θ + 2 cos θ)` = ______.
