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प्रश्न
If cot θ(1 + sin θ) = 4m and cot θ(1 – sin θ) = 4n then prove that (m2 – n2)2 = m
उत्तर
Given `cot theta(1 + sin theta)` = 4 m
`cot theta(1 - sin theta)` = 4 n
(4m)2 – (4n)2 = `cot^2theta(1 + sintheta)^2 - cot^2theta(1 - sintheta)^2`
16m2 – 16n2 = `cot^2theta[(1 + sintheta)^2 - (1 - sin theta)^2]`
16(m2 – n2) = `cot2theta [(1 + 2sintheta + sin^2theta) - (1 - 2sintheta + sin^2theta)]`
16(m2 – n2) = `cot^2theta [1 + 2sintheta + sin^2theta - 1 + 2sintheta - sin^2theta]`
16(m2 – n2) = `cot^2theta*4sintheta`
4(m2 – n2) = `cot^2theta sintheta`
Squaring on both sides
16(m2 – n2)2 = `cot^4theta sin^2theta`
16(m2 – n2)2 = `cot^2theta*cot^2theta sin^2theta`
16(m2 – n2)2 = `cot^2theta("cosec"^2theta - 1) sin^2theta`
16(m2 – n2)2 = `cot^2theta ("cosec"^2theta sin^2theta - sin^2theta)`
16(m2 – n2)2 = `cot^2theta(1 - sin^2theta)`
16(m2 – n2)2 = `cot^2theta cos^2theta` ......(1)
4m . 4n = `cot theta(1 + sin theta) cottheta(1 - sin theta)`
16 mn = `cot^2theta(1 - sin^2theta)`
16 mn = `cot^2theta cos^2theta` ......(2)
Equations (1) and (2), we have
16(m2 – n2)2 = 16 mn
(m2 – n2)2 = mn
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