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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 12 of 160

Marks: +1, -0

\alpha, \beta

are the roots of the equation

\sin^2 x + b \sin x + c = 0

\alpha + \beta = \frac{\pi}{2}

b^2 - 1 =

$C$
$2c$
$C^2$
$4c^2$

Solution:

Given,

\sin^2 x + b \sin x + c = 0

and

\alpha + \beta = \frac{\pi}{2}

Let

y = \sin x

, thus we get

y^2 + b y + c = 0

The roots of this equation are

\sin \alpha

and

\sin \beta

.

So, sum of the roots

= \sin \alpha + \sin \beta = - \frac{b}{1}

Product of the roots

= \sin \alpha \cdot \sin \beta = c

Since,

\alpha + \beta = \frac{\pi}{2}

\Rightarrow \; \; \beta = \frac{\pi}{2} - \alpha

So,

\sin \alpha + \sin \left( \frac{\pi}{2} - \alpha \right) = -b

\; \Rightarrow \sin \alpha + \cos \alpha = -b

\; \Rightarrow ( \because \sin \alpha + \cos \alpha )^2 = (-b)^2

\; \Rightarrow \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = (-b)^2

\; \Rightarrow 1 + 2c = b^2 [ \because \sin^2 \theta + \cos^2 \theta = 1 ]

\; \Rightarrow b^2 - 1 = 2c

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