Advertisements
Advertisements
Question
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Advertisements
Solution
We need to prove cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Solving the L.H.S, we get
`cosec^6 theta = (cosec^2 theta)^3`
`= (1 + cot^2 theta)^3` .......`(1 + cot^2 theta = cosec^2 theta)`
Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2` we get
`(1 + cot^2 theta)^3 = 1 + cot^6 theta + 3(1)^2 (cot^2 theta) + 3(1) (cot^2 theta)^2`
`= 1 + cot^6 theta + 3 cot^2 theta + 3 cot^4 theta`
`= 1 + cot^6 theta + 3 cot^2 theta (1 + cot^2 theta)`
`= 1 + cot^6 theta + 3 cot^2 theta cosec^2 theta` `(using 1 + cot^2 theta = cosec^2 theta)`
Hence proved.
APPEARS IN
RELATED QUESTIONS
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
`(cosecA - 1)/(cosecA + 1) = (cosA/(1 + sinA))^2`
`sin^2 theta + 1/((1+tan^2 theta))=1`
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
sec2θ – tan2θ = ?
`(1 + cot^2A)/(1 + tan^2A)` = ?
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
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`
