Question

e v a l u a t e : question 188. part ii)
thank you​

Answer 1

$\huge\mathfrak{\red{Answer}}$

2

Step-by-step explanation:

$\frac{ \sec(42) }{ \csc(48) } + \frac{3 \tan(50) }{ \cot(40) } - \frac{2 \cos(43) }{ \sin(47) }$

we know that

$\sec(a) and \csc(a) form \: complimentary \: angles$

similarly, cosA and sinA form complimentary angles.

&

tanA and cotA form complimentary angles.

Therefore,

$= \frac{ \sec(90 - 42) }{ \cos(48) } - \frac{3( \tan(90 - 50) }{ \cot(40) } - \frac{2( \cos(90 - 43) ) }{ \sin(47) }$

$= \frac{ \csc(48) }{ \csc(48) } + \frac{3( \cot(40) }{ \cot(40) } - \frac{2( \sin(47) }{47}$

$= 1 + 3 - 2$

$= 2$