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CBSE Class 12-science Answered

differentiate tan-1(1-x2)1/2/ x  wrt cos-1[2x(1-x2)1/2] when x ≠ 0 explain in great detail
Asked by haroonrashidgkp | 13 Aug, 2018, 08:56: PM
answered-by-expert Expert Answer
begin mathsize 16px style Let straight u equals tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 minus straight x squared end root over denominator straight x end fraction close parentheses straight x equals cosθ rightwards double arrow straight theta equals cos to the power of negative 1 end exponent straight x straight u equals tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 minus cos squared straight theta end root over denominator cosθ end fraction close parentheses straight u equals tan to the power of negative 1 end exponent open parentheses sinθ over cosθ close parentheses straight u equals tan to the power of negative 1 end exponent tanθ straight u equals straight theta straight u equals cos to the power of negative 1 end exponent straight x du over dx equals fraction numerator negative 1 over denominator square root of 1 minus straight x squared end root end fraction.......... left parenthesis straight i right parenthesis straight v equals cos to the power of negative 1 end exponent open parentheses 2 straight x square root of 1 minus straight x squared end root close parentheses straight x equals cosθ rightwards double arrow straight theta equals cos to the power of negative 1 end exponent straight x straight v equals cos to the power of negative 1 end exponent open parentheses 2 cosθ square root of 1 minus cos squared straight theta end root close parentheses straight v equals cos to the power of negative 1 end exponent open parentheses 2 sinθcosθ close parentheses straight v equals cos to the power of negative 1 end exponent open parentheses sin 2 straight theta close parentheses straight v equals cos to the power of negative 1 end exponent open square brackets cos open parentheses straight pi over 2 minus 2 straight theta close parentheses close square brackets straight v equals straight pi over 2 minus 2 straight theta straight v equals straight pi over 2 minus 2 cos to the power of negative 1 end exponent straight x dv over dx equals fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction.............. left parenthesis ii right parenthesis du over dv equals fraction numerator du over dx over denominator dv over dx end fraction space space space space space space space space space space space space space space space space from space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis du over dv equals fraction numerator fraction numerator negative 1 over denominator square root of 1 minus straight x squared end root end fraction over denominator fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction end fraction equals negative 1 half end style
Answered by Sneha shidid | 14 Aug, 2018, 10:18: AM

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