If x=sec ө-cos ө, y=secnө-cosn ө then show that

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Question

If x=sec ө-cos ө, y=secⁿө-cosⁿ ө then show that $left parenthesis straight x squared plus 4 right parenthesis open parentheses dy over dx close parentheses squared equals straight n squared open parentheses straight y squared plus 4 close parentheses$

Solution

Here space straight x equals sec space straight theta minus cos space straight theta therefore space dx over dθ equals sec space straight theta space tan space straight theta plus sin space straight theta Also space space space straight y equals sec to the power of straight n space straight theta minus cos to the power of straight n space straight theta therefore space dy over dθ equals straight n space sec to the power of straight n minus 1 end exponent straight theta. sec space straight theta space tan space straight theta minus straight n space cos to the power of straight n minus 1 end exponent straight theta. left parenthesis negative sin space straight theta right parenthesis space space space space space space space space space space space space space equals straight n space sec to the power of straight n straight theta space tan space straight theta plus straight n space cos to the power of straight n minus 1 end exponent straight theta space sin space straight theta Now space dy over dx equals fraction numerator begin display style dy over dθ end style over denominator begin display style dx over dθ end style end fraction equals fraction numerator straight n space sec to the power of straight n straight theta plus straight n space cos to the power of straight n minus 1 end exponent straight theta space sin space straight theta over denominator sec space straight theta space tanθ plus sin space straight theta end fraction space space space space space space space space space space space space space space space space space equals fraction numerator straight n space sec to the power of straight n straight theta. begin display style fraction numerator sin space straight theta over denominator cos space straight theta end fraction end style plus straight n space cos to the power of straight n minus 1 end exponent straight theta space sin space straight theta over denominator sec space straight theta plus cos space straight theta end fraction equals fraction numerator straight n space sin space straight theta left parenthesis sec to the power of straight n space straight theta plus cos to the power of straight n space straight theta right parenthesis over denominator sin space straight theta left parenthesis sec space straight theta plus cos space straight theta right parenthesis end fraction therefore space dy over dx equals fraction numerator straight n left parenthesis sec to the power of straight n space straight theta plus cos to the power of straight n space straight theta right parenthesis over denominator sec space straight theta plus cos space straight theta end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

straight L. straight H. straight S equals left parenthesis straight x squared plus 4 right parenthesis open parentheses dy over dx close parentheses squared space space space space space space space space space space space equals left square bracket left parenthesis sec space straight theta minus cos space straight theta right parenthesis squared equals 4 right square bracket. fraction numerator straight n squared left parenthesis sec to the power of straight n space straight theta plus cos to the power of straight n space straight theta right parenthesis squared over denominator open parentheses sec space straight theta plus cos space straight theta close parentheses squared end fraction space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket space space space space space space space space space space space equals straight n squared open parentheses sec to the power of straight n space straight theta plus cos to the power of straight n space straight theta close parentheses squared space space space space space space space space space space space equals straight n squared left square bracket open parentheses sec to the power of straight n space straight theta minus cos to the power of straight n space straight theta close parentheses squared plus 4 right square bracket equals straight n squared left parenthesis straight y squared plus 4 right parenthesis space space space space space space space space space space space equals straight R. straight H. straight S

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Continuity And Differentiability

If x=sec ө-cos ө, y=secⁿө-cosⁿ ө then show that $left parenthesis straight x squared plus 4 right parenthesis open parentheses dy over dx close parentheses squared equals straight n squared open parentheses straight y squared plus 4 close parentheses$

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