If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' is
- a)x2 + y2 + 2x cot 2α = 1
- b)(cot 2α) (x2 + y2) = 1 + x
- c)x2 + y2+ 2y tan 2α = 1
- d)x2 + y2 + 2x sin 2α = 1
Correct answer is option 'A'. Can you explain this answer?
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This discussion on If α + i β = tan1 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? is done on EduRev Study Group by JEE Students. The Questions and
Answers of If α + i β = tan1 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? are solved by group of students and teacher of JEE, which is also the largest student
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If α + i β = tan1 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? over here on EduRev! Apart from being the largest JEE community, EduRev has the largest solved
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This discussion on If α + i β = tan1 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? is done on EduRev Study Group by JEE Students. The Questions and
Answers of If α + i β = tan1 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? are solved by group of students and teacher of JEE, which is also the largest student
community of JEE. If the answer is not available please wait for a while and a community member will probably answer this
soon. You can study other questions, MCQs, videos and tests for JEE on EduRev and even discuss your questions like
If α + i β = tan1 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? over here on EduRev! Apart from being the largest JEE community, EduRev has the largest solved
Question bank for JEE.
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