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Please somebody answer from question 11...I am too tired to do it on my own...please somebody​

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Answer:

\sin(a + b) = 1 \\( a + b) = \sin {}^{ - 1} (1) = 90 \\ (a + b) = 90.......(1) \\ \cos(a - b) = 1 \\ (a - b) = \cos {}^{ - 1} (1) = 180 \\ (a - b) = 180.......(2) \\ from \: solving \: of\: (1) \: and \: (2) \: we \: will \: get\\ 2a = 270 \: \: = > a = 135 \\ thus \: \: b = - 45 \\ thus \: its \: follows \: the \: <klux>GIVEN</klux> \: condition \: as \\0 < ( a + b ) \leqslant 90 \\ and \: a > b \\ so \: the \: answer \: is \\ a \: = 135 \\ b = - 45

similar METHOD for question no. 11

for question no 12 put the value as sin(45 - 30) and cos(45 - 30) instead of 15 where a = 45 and b = 30 and FOLLOW the method as given in question. now put the value of all quantities and try to simplify you will get your answer


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