Solve the following equations  2 tan to the power of negative 1 space end exponent space left parenthesis cos space x right parenthesis equals space tan to the power of negative 1 end exponent space left parenthesis 2 space cos e c space x right parenthesis

Asked by Anshul | 26th May, 2017, 06:14: PM

Expert Answer:

begin mathsize 16px style 2 space tan to the power of negative 1 end exponent left parenthesis cosx right parenthesis space equals space tan to the power of negative 1 end exponent left parenthesis 2 cosecx right parenthesis
rightwards double arrow tan to the power of negative 1 end exponent open parentheses fraction numerator 2 cosx over denominator 1 minus cos squared straight x end fraction close parentheses equals space tan to the power of negative 1 end exponent left parenthesis 2 cosecx right parenthesis space..... open square brackets Since space 2 tan to the power of negative 1 end exponent straight x equals open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses close square brackets
rightwards double arrow fraction numerator 2 cosx over denominator sin squared straight x end fraction equals space 2 over sinx
rightwards double arrow sinx equals space cosx
rightwards double arrow straight x equals straight pi over 4
end style

Answered by Rebecca Fernandes | 24th Nov, 2017, 12:21: PM

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