Question

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Solve this!!
Q. Given 5 cosA - 12 sinA = 0, find the value of sinA + cosA / 2 cosA - sinA.

Answer 1

5cosA-12sinA=0

5cosA = 12sinA

sinA = 5cosA /12   .... (1)

(sinA + cosA)   .... (2)

(2cosA -sinA)

sub (1) in (2)

 (5cosA/12) + cosA

2cosA- (5cosA/12)

LCM= 12

(5cosA + 12 cosA) / 12

(24cosA- 5cosA) /12

12 get cancelled

5cosA +12 cosA

24cosA – 5cosA

17 cosA

19cosA

cosA get cancelled

Therefore the result is 17/19

Answer 2

Given 5cosA - 12sinA = 0

$5 - \frac{12sinA}{cosA} = 0$

5 - 12tanA = 0

-12tanA = -5

tanA = 5/12.


Now,

We know that sec^2A = 1 + tan^2A

                        sec^2A = 1 + (5/12)^2

                         sec^2A = 1 + 25/144

                         sec^2A = 169/144

                         secA = 13/12.


Now,

We know that secA = 1/cosA

                         13/12 = 1/cosA

                         13cosA = 12
         
                           cosA = 12/13..



Now,

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

                        sin^2A + (12/13)^2 = 1

                       sin^2A + 144/169 = 1

                       sin^2A = 1 - 144/169

                       sin^2A = 169 -144/169

                       sin^2A = 25/169

                        sinA = 5/13.



Therefore :


$\frac{sinA + cosA}{2cosA - sinA} = \frac{ \frac{5}{13} + \frac{12}{13} }{2( \frac{12}{13}) - \frac{5}{13} }$


$= \frac{17}{13} * \frac{13}{19}$


$= \frac{17}{19}$



Hope this helps!