Question

Hola!!!

Help me out with this question.

Thanx

Answer 1

Note: I am writing theta as A.

Given SecA = (13/5).

(1) We know that cosA = 1/secA

= > cosA = (1/13/5)

                = 5/13.


(2) We know that sin^2A + cos^2A = 1

= > sin^2A = 1 - cos^2A

= > sin^2A = 1 - (5/13)^2

= > sin^2A = 1 - 25/169

= > sin^2A = 144/169

= > sin = 12/13  


Now,

$= \ \textgreater \ \frac{2sinA - 3cosA}{4sinA - 9cosA}$

$= \ \textgreater \ \frac{2( \frac{12}{13}) - 3( \frac{5}{13}) }{4( \frac{12}{13}) - 9( \frac{5}{13} ) }$

$= \ \textgreater \ \frac{ \frac{24}{13} - \frac{15}{13} }{ \frac{48}{13} - \frac{45}{13} }$

$= \ \textgreater \ \frac{ \frac{9}{13} }{ \frac{3}{13} }$

$= \ \textgreater \ \frac{9}{3}$

= > 3.


Therefore the final answer is 3.



Hope this helps!

Answer 2

The answer is in the attachment.Hope you like it.