Advertisements
Advertisements
Question
sin4A – cos4A = 1 – 2cos2A हे सिद्ध करा.
Solution
डावी बाजू = sin4A – cos4A
= (sin2A)2 – (cos2A)2
= (sin2A + cos2A)(sin2A – cos2A) .....[∵ a2 – b2 = (a + b)(a – b)]
= (1)(sin2A – cos2A) ......[∵ sin2A + cos2A = 1]
= sin2A – cos2A
= (1 – cos2A) – cos2A ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"" = sin^2"A")]`
= 1 – 2cos2A
= उजवी बाजू
∴ sin4A – cos4A = 1 – 2cos2A
APPEARS IN
RELATED QUESTIONS
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
cot θ + tan θ = cosec θ sec θ
`1/(secθ - tanθ)` = secθ + tanθ
sec4θ - cos4θ = 1 - 2cos2θ
cot2θ - tan2θ = cosec2θ - sec2θ
cot2θ × sec2θ = cot2θ + 1 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
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
