Question

cot35/tan55 -1/2[sec23/cosec67]+3sin41 sec49

Answer 1

$\frac{ \cot(35) }{ \tan(55) } - \frac{1}{2} ( \frac{ \sec(23) }{ \csc(67) }) + 3( \sin(41))( \sec(49 ) \\ = \frac{ \cot(35) }{ \cot(35) } - \frac{1}{2} \frac{ \sec(23) }{ \sec(23) } + 3( \frac{ \sin(41) }{ \cos(49) } ) \\ = 1 - ( \frac{1}{2} ) + 3( \frac{ \sin(41) }{ \sin(41) } ) \\ = \frac{1}{2} + 3(1) \\ = 3 \frac{1}{3 } = \frac{10}{3}$

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