Advertisements
Advertisements
Question
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
Advertisements
Solution
L.H.S = sec4 θ (1 – sin4 θ) – 2 tan2 θ
= `1/cos^4 theta [1 - (sin^2 theta)^2]- 2 xx (sin^2 theta)/(cos^2 theta)`
= `1/(cos^4 theta) (1 + sin^2 theta) (1 - sin^2 theta) - 2 (sin^2 theta)/(cos^2 theta)`
= `1/(cos^4 theta) xx cos^2 theta (1 + sin^2 theta) - 2 (sin^2 theta)/(cos^2 theta)`
= `(1 + sin^2 theta)/(cos^2 theta) - (2sin^2 theta)/(cos^2 theta)`
= `(1 + sin^2 theta - 2sin^2 theta)/(cos^2 theta)`
= `(1 - sin^2 theta)/(cos^2 theta)`
= `(cos^2 theta)/(cos^2 theta)`
L.H.S = R.H.S
∴ sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
APPEARS IN
RELATED QUESTIONS
As observed from the top of an 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse of the horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.
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 sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is ______.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
If tan θ = `x/y`, then cos θ is equal to ______.
