Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
उत्तर
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
संबंधित प्रश्न
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
Simplify : 2 sin30 + 3 tan45.
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Choose the correct alternative:
cot θ . tan θ = ?
sin2θ + sin2(90 – θ) = ?
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
If 2sin2θ – cos2θ = 2, then find the value of θ.
(1 – cos2 A) is equal to ______.
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`
