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Question
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
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Solution
डावी बाजू = sec2A – cosec2A
= `1/(cos^2"A") - 1/(sin^2"A")`
= `(sin^2"A" - cos^2"A")/(cos^2"A"*sin^2"A")`
= `(sin^2"A" - (1 - sin^2"A"))/(sin^2"A"*cos^2"A")` .....`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 sin^2"A" = cos^2"A")]`
= `(sin^2"A" - 1 + sin^2"A")/(sin^2"A"*cos^2"A")`
= `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
= उजवी बाजू
∴ sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
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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)`
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= उजवी बाजू
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
दाखवा की: `tanA/(1 + tan^2 A)^2 + cotA/(1 + cot^2A)^2` = sinA × cosA.
sin2θ + cos2θ ची किंमत काढा.
उकलः
Δ ABC मध्ये, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` ...(पायथागोरसचे प्रमेय)
दोन्ही बाजूला AC2 ने भागून,
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
परंतु `"AB"/"AC" = square "आणि" "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
