Answers

  • Brainly User
2014-11-12T16:07:02+05:30
3+1/2+5/2=12/2=6
as tan(67) can be written tan(90-23)=cot23.....1
sin42=sin(90-48)=cos48.............2
cosec61=cosec(90-29)=sec29...............3
so they can cancel each other out
and we left with
3+1/2+5/2=
12/2=
6


0
your answer is wrong so for saying this
Comment has been deleted
2014-11-13T16:55:40+05:30

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You should use parentheses when writing algebraic expressions.  It is not clear sometimes, which factor is in numerator and which is in denominator.  The answer can be one of the following.

3\frac{ Tan\ 67}{Cot\ 23}+\frac{1}{\frac{2\ Sin\ 42}{Cos\ 48}}+\frac{5}{\frac{2\ Cosec\ 61}{Sec\ 29}}\\\\=3\frac{tan\ 67}{Cot(90-67)}+\frac{1}{2}\frac{Cos48}{Sin42}+\frac{5}{2}\frac{Sec29}{Cosec61}\\\\=3\frac{tan67}{tan67}+\frac{1}{2}\frac{Cos58}{Sin(90-58)}+\frac{5}{2}\frac{sin61}{cos29}=3+\frac{cos58}{cos58}+\frac{5}{2}\frac{sin61}{Cos(90-61)}\\\\=3+1/2+5/2=6\\\\

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3\frac{ Tan\ 67}{Cot\ 23}+\frac{1}{2\ Sin\ 42\ *\ Cos\ 48}+\frac{5}{2\ Cosec\ 61\ *\ Sec\ 29}\\\\=3\frac{ Tan\ 67}{Cot\ (90-67)}+\frac{1}{2\ Sin\ 42\ *\ Cos(90-42)}+\frac{5}{2}Sin\ 61\ *\ Cos\ 29\\\\=3\frac{Tan\ 67}{Tan\ 67}+\frac{1}{2\ Sin^2\ 42}+\frac{5}{2}Sin\ 61*Sin61\\\\=3+3.029=6.029\\
0
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