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Question
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
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Solution
L.H.S. = sin4θ - cos4θ
L.H.S. = (sin2θ)2 - (cos2θ)2
L.H.S. = (sin2θ - cos2θ)(sin2θ + cos2θ)
L.H.S. = (sin2θ - cos2θ) x 1
L.H.S. = sin2θ - cos2θ
L.H.S. = R.H.S.
L.H.S.= sin2θ - (1 - sin2θ)
L.H.S. = sin2θ - 1 + sin2θ
L.H.S. = 2sin2θ - 1
L.H.S. = R.H.S
L.H.S. = 2(1 - cos2θ) - 1
L.H.S. = 2 - 2cos2θ - 1
L.H.S. = 1 - 2cos2θ
L.H.S. = R.H.S.
RELATED QUESTIONS
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
Prove that `sqrt((1 + cos theta)/(1 - cos theta)) + sqrt((1 - cos theta)/(1 + cos theta)) = 2 cosec theta`
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Choose the correct alternative:
sec2θ – tan2θ =?
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
