Question

3 tan67⁰÷cot23⁰₊1÷2 sin42⁰÷cos48⁰₊5÷2 cosec61⁰÷sec29⁰

Answer 1

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

Answer 2

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