Math, asked by rekhaahlawat624, 11 months ago

Please please answer this please

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Answered by ajayshotia

put x=sec@ then@=sec¹-x
xsin@/1-xcos@=sec@sin@/1-sec@cos@
=(sin@/cos@)/(1-cos@/cos@)
=tan@/(1-1)=tan@/0=infinity
tan-¹(infinity)=tan-¹(tanπ/2)=π/2

rekhaahlawat624: thanks but sorry answer is something else ...ur is also right but u know derivatives can be found in different ways..

ajayshotia: yes but can u tell me ans

ajayshotia: oh derivative karna tha iska

ajayshotia: sorry

ajayshotia: y=tan-¹(xsin@/1-xcos@) ,or tany=xsin@/1-xcos@,d/dx tany=d/dx (xsin@/1-xcos@), sec²ydy/dx=1-xcos@)(d/dxsin@)-[xsin@d/dx(1-xcos@)]/[(1-xcos@)²], sec²ydy/dx=(1-xcos@)sin@-xsin@(-cos@)/(1-xcos@)², dy/dx=1/sec²y[sin@-xcos@sin@+xsin@cos@]/(1-xcos@)², dy/dx=cos²y[sin²@/(1-xcos@)²], u can use y =tan-¹(xsin@/1-xcos@) in place of y in terms cos²y

rekhaahlawat624: hmm thnku soooo mch

ajayshotia: aapki samjh me aa gya ye

rekhaahlawat624: not completely but method i m getting

ajayshotia: ok in which class

rekhaahlawat624: 12th class

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