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Question
`sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ हे सिद्ध करा.
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Solution
डावी बाजू = `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)`
= `sintheta/(1/costheta + 1) + sintheta/(1/costheta - 1`
= `sintheta/((1 + costheta)/costheta) + sintheta/((1 - costheta)/(costheta))`
= `(sintheta costheta)/(1 + costheta) + (sintheta costheta)/(1 - costheta)`
= `sin theta costheta (1 /(1 + costheta) + 1/(1 - costheta))`
= `sintheta costheta [(1 - costheta + 1 + costheta)/((1 + costheta)(1 - costheta))]`
= `sintheta costheta (2/(1 - cos^2theta))` ......[∵ (a + b)(a – b) = a2 – b2]
= `sintheta costheta xx 2/(sin^2theta)` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
= `2 xx (costheta)/(sintheta)`
= 2cot θ
= उजवी बाजू
∴ `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
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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`
