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NDA Mathematics Solved Question Paper 2019

Multiple Choice Questions

What is the value of $\frac{\sin (34 °) \cos (236 °) - \sin (56 °) \sin (124 °)}{\cos (28 °) \cos (88 °) + \cos (178 °) \sin (208 °)}$ = ?

$\tan (54 °) can be expressed as$

$\frac{\sin (9 °) + \cos (9 °)}{\sin (9 °) - \cos (9 °)}$
$\frac{\sin (9 °) - \cos (9 °)}{\sin (9 °) + \cos (9 °)}$
$\frac{\cos (9 °) + \sin (9 °)}{\cos (9 °) - \sin (9 °)}$
$\frac{\sin (36 °)}{\cos (36 °)}$

If p = $Xcos (θ) - Ysin (θ)$ , q = $Xsin (θ) + Ycos (θ)$ and p² + 4pq + q² = AX² + BY², $0 \leq θ \leq \frac{π}{2}$ . What is the value of $θ$ ?

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{6}$

$If p = (Xcos (θ) - Ysin (θ)), q = Xsin (θ) + Ycos (θ) and p^{2} + 4 pq + q^{2} = {AX}^{2} + {BY}^{2}, 0 \leq θ \leq \frac{π}{2}$

What is the value of A ?

5.
$If p = (Xcos (θ) - Ysin (θ)), q = Xsin (θ) + Ycos (θ) and p^{2} + 4 pq + q^{2} = {AX}^{2} + {BY}^{2}, 0 \leq θ \leq \frac{π}{2}$

-1

0

1

2

-1

$p = X \cos (θ) - Y \sin (θ) q = X \sin (θ) + Y \cos (θ) p^{2} = X^{2} \cos^{2} (θ) + Y^{2} \sin^{2} (θ) - 2 XYsin (θ) \cos (θ) q^{2} = X^{2} \sin^{2} (θ) + Y^{2} \cos^{2} (θ) + 2 XYsin (θ) \cos (θ) p^{2} + q^{2} = X^{2} (\sin^{2} (θ) + \cos^{2} (θ)) + Y^{2} (\sin^{2} (θ) + \cos^{2} (θ)) p^{2} + q^{2} = X^{2} + Y^{2} pq = (Xcos (θ)) (Ysin (θ)) + (Xcos (θ)) (Ycos (θ)) - (Ysin (θ)) (Xsin (θ)) - (Y \sin (θ)) (Ycos (θ)) pq = X^{2} \sin (θ) \cos (θ) + {XYcos}^{2} (θ) - XY \sin^{2} (θ) - Y^{2} \sin (θ) \cos (θ) 4 pq = 2 X^{2} \sin (2 θ) - 2 Y^{2} \sin (2 θ) + 4 XYcos (2 θ) p^{2} + q^{2} + 4 pq = X^{2} (1 + 2 \sin (2 θ)) Compare with {AX}^{2} + {BY}^{2} We get, A = 1 + 2 \sin (2 θ) B = 1 - 2 \sin (2 θ) 4 XYcos (2 θ) = 0 ∵ B = 1 - 2 \sin (2 θ) ∵ θ = \frac{π}{4} ∴ B = 1 - 2 \sin (\frac{π}{2}) = 1 - 2 = - 1$