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Question
(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
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Solution
डावी बाजू = (sec θ - cos θ)(cot θ + tan θ)
= `(1/cos θ - cos θ) (cos θ/sin θ + sin θ/cos θ)`
= `((1 - cos^2θ)/cos θ)((cos^2θ + sin^2θ)/(sinθcosθ))`
= `sin^2θ/cosθ xx 1/(sinθcosθ)` ....`[(∵ sin^2θ + cos^2θ = 1),(∴ sin^2θ = 1 - cos^2θ)]`
= `sinθ/cosθ . 1/cosθ`
= tan θ . sec θ
= उजवी बाजू
∴ (sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ
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sec6x - tan6x = 1 + 3sec2x × tan2x
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 cos θ = `24/25`, तर sin θ = ?
sec2θ − cos2θ = tan2θ + sin2θ हे सिद्ध करा.
cot2θ – tan2θ = cosec2θ – sec2θ हे सिद्ध करा.
2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0 हे सिद्ध करा.
जर tan θ – sin2θ = cos2θ, तर sin2θ = `1/2` हे दाखवा.
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
