Advertisements
Advertisements
प्रश्न
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Advertisements
उत्तर
L.H.S. = sin2 θ + cos4 θ
= 1 - cos2 θ + cos4 θ
= 1 - cos2 θ (1 - cos2 θ)
= 1 - (1 - sin2 θ) sin2 θ
= 1 - sin2 θ + sin4 θ
= cos2 θ + sin4 θ
= R.H.S.
Hence proved.
संबंधित प्रश्न
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
If tanθ `= 3/4` then find the value of secθ.
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
