Advertisements
Advertisements
Question
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
Options
1
`1/2`
2
3
Advertisements
Solution
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is 1.
Explanation:
Given,
sinA + sin2A = 1
⇒ sinA = 1 – sin2A = cos2A ...[∵ sin2θ + cos2θ = 1]
On squaring both sides, we get
sin2A = cos4A
⇒ 1 – cos2A = cos4A
⇒ cos2A + cos4A = 1
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
If cos \[9\theta\] = sin \[\theta\] and \[9\theta\] < 900 , then the value of tan \[6 \theta\] is
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
If cosθ = `5/13`, then find sinθ.
Choose the correct alternative:
1 + cot2θ = ?
If tan θ = `13/12`, then cot θ = ?
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
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`
