Advertisements
Advertisements
Question
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
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
APPEARS IN
RELATED QUESTIONS
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 identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Choose the correct alternative:
1 + tan2 θ = ?
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
(sec θ + tan θ) . (sec θ – tan θ) = ?
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
If tan α + cot α = 2, then tan20α + cot20α = ______.
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
