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If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer? for JEE 2022 is part of JEE preparation. The Question and answers have been prepared
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If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer?, a detailed solution for If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.