In a ABC, A and B satisfy c² x²-c(a+b) x+a b=0 then

Correct Answer is : sin⁡A+cos⁡A=a+bc A + A = a+b/csinA+cosA=ca+b

Correct Answer is : sin⁡A+cos⁡A=a+bc A + A = a+b/csinA+cosA=ca+b

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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 26 of 160

Marks: +1, -0

In a

\triangle ABC, \sin A

\sin B

c^{2} x^{2}-c(a+b) x+a b=0

then

Solution:

Consider the equation,

c^{2} x^{2}-c(a+b) x+a b=0

c^{2} x^{2}-c a x-c b x+a b=0

(c x-a)(c x-b)=0

x=\frac{a}{c}, \frac{b}{c}

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

\sin A = \frac{a}{c}

c = \frac{a}{\sin A}

And,

\sin B = \frac{b}{c}

c = \frac{b}{\sin B}

From the sine law,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Then,

c \sin c = c

\sin c = 1

c = \frac{\pi}{2}

Then in right angled triangle

ABC

\sin A = \frac{a}{c}

and

\cos A = \frac{b}{c}

So,

\sin A + \cos A = \frac{a}{c} + \frac{b}{c}

= \frac{a+b}{c}

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