Which is the correct option? Let 
begin mathsize 20px style P equals open curly brackets theta colon sin theta minus cos theta equals square root of 2 cos theta close curly brackets space a n d space Q equals open curly brackets theta colon sin theta plus cos theta equals square root of 2 sin theta close curly brackets
b e space t w o space s e t s. T h e n
open parentheses a close parentheses P subset of Q space a n d space Q minus P space n o t space e q u a l space t o empty set
open parentheses b close parentheses Q ⊄ P
open parentheses c close parentheses P ⊄ Q
open parentheses d close parentheses P equals Q end style

Asked by sunil2791 | 11th Jun, 2017, 11:31: AM

Expert Answer:

begin mathsize 16px style straight P equals left curly bracket straight theta colon sinθ minus cosθ equals square root of 2 cosθ right curly bracket
straight Q equals left curly bracket straight theta colon sinθ plus cosθ equals square root of 2 sinθ right curly bracket
Consider comma space
sinθ minus cosθ equals square root of 2 cosθ
rightwards double arrow cos open parentheses straight pi over 2 minus straight theta close parentheses minus cosθ equals square root of 2 cosθ
rightwards double arrow 2 sin open parentheses fraction numerator straight pi over 2 minus straight theta plus straight theta over denominator 2 end fraction close parentheses sin open parentheses fraction numerator straight theta minus straight pi over 2 plus straight theta over denominator 2 end fraction close parentheses equals square root of 2 cosθ
rightwards double arrow 2 sin straight pi over 4 sin open parentheses straight theta minus straight pi over 4 close parentheses equals square root of 2 cosθ
rightwards double arrow 2 cross times fraction numerator 1 over denominator square root of 2 end fraction sin open parentheses straight theta minus straight pi over 4 close parentheses equals square root of 2 cosθ
rightwards double arrow sin open parentheses straight theta minus straight pi over 4 close parentheses equals cosθ
rightwards double arrow sin open parentheses straight theta minus straight pi over 4 close parentheses equals sin open parentheses straight pi over 2 minus straight theta close parentheses
rightwards double arrow straight theta minus straight pi over 4 equals straight pi over 2 minus straight theta
rightwards double arrow straight theta equals fraction numerator 3 straight pi over denominator 8 end fraction.......................... left parenthesis straight i right parenthesis
Similarly comma
sinθ plus cosθ equals square root of 2 sinθ
sinθ plus sin open parentheses straight pi over 2 minus straight theta close parentheses equals square root of 2 sinθ
2 sin open parentheses fraction numerator straight theta plus straight pi over 2 minus straight theta over denominator 2 end fraction close parentheses cos open parentheses fraction numerator straight theta minus straight pi over 2 plus straight theta over denominator 2 end fraction close parentheses equals square root of 2 sinθ
2 sin straight pi over 4 cos open parentheses straight theta minus straight pi over 4 close parentheses equals square root of 2 sinθ
2 cross times fraction numerator 1 over denominator square root of 2 end fraction cos open parentheses straight theta minus straight pi over 4 close parentheses equals square root of 2 sinθ
cos open parentheses straight theta minus straight pi over 4 close parentheses equals cos open parentheses straight pi over 2 minus straight theta close parentheses
straight theta minus straight pi over 4 equals straight pi over 2 minus straight theta
2 straight theta equals straight pi over 2 plus straight pi over 4
straight theta equals fraction numerator 3 straight pi over denominator 8 end fraction....................... left parenthesis ii right parenthesis
From space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
straight P equals straight Q
end style

Answered by Sneha shidid | 12th Jun, 2017, 09:30: AM