Question

Good evening friends!

Evaluate :-

(i)
$\frac{ \sin(18 {}^{0} ) }{ \cos(72 {}^{0} ) }$

(ii)
$\cos48 {}^{0} - \sin59 {}^{0}$
Please explain the answer.

Thanks!

Answer 1

Here is the solution of ques. 1

Answer 2

Hi sis!


Here is yr answer......
$1. \\ = > \frac{ \sin18 }{ \cos72} \\ \\ = > \frac{ \sin18}{ \sin(90 - 72) } \: \: \: ( \ \cos(90 - \alpha ) = \sin \alpha \\ \\ = > \frac{ \sin18 }{ \sin18 } \\ \\ = > \frac{1}{1} \\ \\ = > 1 \\ \\ \\ \\ \\ 2. \\ = > \ \cos 48 - \sin59 \\ \\ = > \cos48 - \cos(90 - 59) \\ \\ = > \cos48 - \cos31 \\ \\ = > \cos17$



Hope it hlpz....