TS EAMCET 4 May 2018 Shift 2 Solved Paper

Consider the equation,
c2x2−c(a+b)x+ab=0
c2x2−cax−cbx+ab=0
(cx−a)(cx−b)=0
x=

a

c

,

b

c

Since, Sin A and Sin B are the roots of the equation so,
sin‌A=

a

c

c=

a

sin‌A

And,
sin‌B=

b

c

c=

b

sin‌B

From the sine law,

a

sin‌A

=

b

sin‌B

=

c

sin‌C

Then,
c‌sin‌c=c
sin‌c=1
c=

π

2

Then in right angled triangle ABC,
sin‌A=

a

c

and cos‌A=

b

c

So,
sin‌A+cos‌A=

a

c

+

b

c

=

a+b

c