​$$\text{If } A \text{ lies in the second quadrant and } 13\sin A - 12 = 0, \text{ then } 5\sin A + 12\cos A + \tan^2 A - (\sec^2 A - 1) = ?$$​​

Correct option is A

Given:

​$$13\sin A - 12 = 0$$ 

$$5\sin A + 12\cos A + \tan^2 A - (\sec^2 A - 1)= ?$$ 

Concept Used:

​$$\sin^2 A + \cos^2 A = 1$$ 

$$\tan A = \frac{\sin A}{\cos A}$$ 

$$\sec^2A-1= \tan^2A$$ 

Solution: ​

= $$13\sin A - 12 = 0$$ 

$$\Rightarrow 13\sin A = 12$$ 

$$\Rightarrow \sin A = \frac{12}{13}$$ 

$$\text{Using the Pythagorean identity }, \\ \ \\ \sin^2 A + \cos^2 A = 1,$$ 

$$\left(\frac{12}{13}\right)^2 + \cos^2 A = 1 \\ \ \\ \Rightarrow \frac{144}{169} + \cos^2 A = 1 \\ \ \\ \Rightarrow \cos^2 A = \frac{25}{169}$$ 

​$$\text{Since A lies in the second quadrant}, ( cos A = -\frac{5}{13})$$ 

= $$5\sin A + 12\cos A + \tan^2 A - (\sec^2 A - 1)$$ 

= $$5\sin A + 12\cos A + \tan^2 A - \tan^2 A$$ 

= $$5\sin A + 12\cos A$$ 

Now, substitute the values of sinA  and  cosA from above 

= $$5 \times \frac{12}{13} + 12 \times \left(-\frac{5}{13}\right)$$ 

= $$\frac{60}{13} - \frac{60}{13} = 0$$ 

​Thus, the value of the expression is 0.

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