Advertisements
Advertisements
Question
sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ हे सिद्ध करा.
Advertisements
Solution
डावी बाजू = sin θ (1 – tan θ) – cos θ (1 – cot θ)
= `sintheta (1 - (sintheta)/(costheta)) - costheta (1 - (costheta)/(sintheta))`
= `sintheta - (sin^2theta)/costheta - costheta + (cos^2theta)/sintheta`
= `sintheta + (cos^2theta)/sintheta - (sin^2theta)/costheta - costheta`
= `(sin^2theta + cos^2theta)/sintheta - ((sin^2theta + cos^2theta)/costheta)`
= `1/sintheta - 1/costheta` ......[∵ sin2θ + cos2θ = 1]
= cosec θ – sec θ
= उजवी बाजू
∴ sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ
APPEARS IN
RELATED QUESTIONS
sec4A(1 - sin4A) - 2tan2A = 1
sec θ(1 - sin θ) (sec θ + tan θ) = 1
cos2θ . (1 + tan2θ) = 1 हे सिद्ध करण्यासाठी खालील कृती पूर्ण करा.
कृती: डावी बाजू = `square`
= `cos^2theta xx square` .........`[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= उजवी बाजू
जर 3 sin θ = 4 cos θ, तर sec θ = ?
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
जर sec θ = `41/40`, तर sin θ, cot θ, cosec θ च्या किमती काढा.
जर cos A = `(2sqrt("m"))/("m" + 1)`, असेल, तर सिद्ध करा cosec A = `("m" + 1)/("m" - 1)`
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
(sin A + cos A) (cosec A – sec A) = cosec A . sec A – 2 tan A हे सिद्ध करा.
θ चे निरसन करा:
जर x = r cosθ आणि y = r sinθ
