Question

If seco = 13/5, find the value of 2sin +3 cos / 5cos 0 - 4 sin :​

Answer 1

Solution

sec∅=13/5

=>cos∅=5/13

now.....

sin∅=√(1-(5/13)²)=12/13

therefore

=(2sin∅ +3 cos∅ )/( 5cos ∅- 4 sin∅ )

=[2(12/13)+3(5/13)]/[5(5/13)-4(12/13)]

=(39/13)/(-23/13)

=-(39/23)

Answer 2

Given

$\sec( \alpha ) = \frac{13}{5}$

We know 1/sec@=cos@ then

$\cos( \alpha ) = \frac{5}{13}$

Also

$\sin( \alpha ) = \sqrt{1 - { \cos( \alpha ) }^{2} } \\ = > \sqrt{1 - {( \frac{5}{13}) }^{2} } \\ = > \sqrt{ \frac{169 - 25}{169} } \\ = > \sqrt{ \frac{144}{169} } \\ = > \frac{12}{13}$

To find

$\frac{2 \sin( \alpha ) + 3 \cos( \alpha ) }{5 \cos( \alpha ) - 4 \sin( \alpha ) }$

put the value of sin@ and cos @

$= > \frac{2( \frac{12}{13}) + 3( \frac{5}{13}) }{5( \frac{5}{13}) - 4( \frac{12}{13}) } \\ = > \frac{ (\frac{24}{13} + \frac{15}{13}) }{( \frac{25}{13} - \frac{48}{13} )} \\ = > \frac{( \frac{39}{13}) }{ ( \frac{ - 23}{13} )}$

13 gets cancel out then,

= > -(39/23)