Advertisements
Advertisements
प्रश्न
sec4θ - cos4θ = 1 - 2cos2θ
Advertisements
उत्तर
डावी बाजू = sec4θ - cos4θ
= (sec2θ)2 – (cos2θ)2
= (sec2θ + cos2θ) (sec2θ – cos2θ) ....[∵ a2 – b2 = (a + b)(a – b)]
`= ((1 + cos^4 theta)(1- cos^4 theta))/(cos^4 theta)`
= `(1 + cos^4 theta)(1- cos^4 theta) (1 + cos^2 theta)/cos^4 theta`
Thus the solution would be not coming equal to RHS.
The correct question would be sin4 θ in place of sec4θ.
On solving this question we get,
= (sin2θ)2 – (cos2θ)2
= (sin2θ + cos2θ) (sin2θ – cos2θ) ....[∵ a2 – b2 = (a + b)(a – b)]
= (1) (sin2θ – cos2θ) ....[∵ sin2θ + cos2θ = 1]
= sin2θ – cos2θ
= (1 - cos2θ) - cos2θ ....[∵ sin2θ = 1 - cos2θ]
= 1 - 2cos2θ = उजवी बाजू
∴ sin4 θ - cos4θ = 1 - 2 cos2θ
APPEARS IN
संबंधित प्रश्न
cot θ + tan θ = cosec θ sec θ
जर secθ = `13/12` , तर इतर त्रिकोणमितीय गुणोत्तरांच्या किमती काढा.
`(cos^2theta)/(sintheta) + sintheta` = cosec θ हे सिद्ध करा.
tan2θ – sin2θ = tan2θ × sin2θ हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= उजवी बाजू
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
sec2θ – cos2θ = tan2θ + sin2θ हे सिद्ध करा.
`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
जर cos A + cos2A = 1, तर sin2A + sin4A = ?
