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If `x=cos e ctheta-sinthetaa n dy=cos e c^ntheta-sin^ntheta,`then show that`(x^2+4)((dy)/(dx))^2=n^2(y^2+4)dot` |
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Answer» As x = cosec theta - sin theta, we have `x^(2)+4=(cosec theta- sin theta)^(2)+4=(cosec theta + sin theta)^(2)" (1)"` `"and "y^(2)+4=(cosec^(n)theta - sin^(n)theta)^(2)+4 = (cosec^(n)theta+ sin^(n)theta)^(2)" (2)"` Now, `(dy)/(dx)=(((dy)/(d""theta)))/(((dx)/(d""theta)))=(n(cosec^(n-1)theta)(-cosec theta cot theta)-n sin^(n-1)theta cos theta)/(-cosec theta cot theta- cos theta)` `=(n(cosec^(n)thetacot theta +sin^(n-1)theta cos theta))/((cosec theta cot theta +cos theta))` `(ncot theta (cosec^(n)theta + sin^(n)theta))/(cot theta (cosec theta +sin theta))` `=(n(cosec^(n)theta+sin^(n)theta))/((cosec theta +sin theta))=(nsqrt(y^(2)+4))/(sqrt(x^(2)+4))" [from (1) and (2) ]"` `"Squaring both sides, we get "((dy)/(dx))^(2)=(n^(2)(y^(2)+4))/((x^(2)+4))` `"or "(x^(2)+4)((dy)/(dx))^(2)=n^(2)(y^(2)+4)` |
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