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प्रश्न
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
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उत्तर
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.
संबंधित प्रश्न
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
Prove the following trigonometric identities:
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
