Question

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

Answer 1

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

= 3 tan 67 / tan (90-23) + 1 / 2 sin 42 / sin (90-48) + 5 / 2 cosec 61 / cosec (90-23)

(SINCE cot theta = tan (90 - theta) ; cos A = sin (90 -A) ; sec B = cosec (90-B)

=[ 3 (tan 67 / tan 67) ]+ [1 / 2 (sin 42 / sin 42) ] + [  5 /  2 (cosec 61 / cosec 61)}

= 3 + 1/2 + 5/2

=3 + 3

=9

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