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Question
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
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Solution
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
⇒ `sin^4A + cos^4A + 2sin^2Acos^2A = 1`
LHS = `(sin^2A + cos^2A)^2`
= 1 = RHS
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RELATED QUESTIONS
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
cos θ . sec θ = ?
Show that tan4θ + tan2θ = sec4θ – sec2θ.
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
