Test Index

NDA Math Trigonometric Ratios of Angles and Equations

Consider the following for the next 03 (three) items:

If p = X cos θ−Y sin θ , q = X sin θ + Y cos θ and p

^{2}

+ 4pq + q

^{2}

= A X

^{2}

+ BY

^{2}

, 0≤ θ ≤

\frac{\pi}{2}

[NDA I 2019]

Question : 2 of 100

Marks: +1, -0

What is the value of θ ?

[NDA I 2019]

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

Validate

Solution:

(C)

Given,

p = Xcosθ − Y sin θ

q = X sin θ + Y cosθ

^{2}

+ 4 pq + q

^{2}

= A X

^{2}

+BY

^{2}

…(i)

^{2}

+ 4 pq + q

^{2}

= X

^{2}

cos

^{2}

θ + Y

^{2}

sin

^{2}

θ − 2XY cosθ sin θ + x

^{2}

sin

^{2}

θ + Y

^{2}

cos

^{2}

θ + 2 XY sin θ cosθ + 4[(Xcosθ− Ysinθ) (X sinθ + Y cosθ) ]

= X

^{2}

+ Y

^{2}

+ 4 [ X

^{2}

sin θ cosθ + XY cos

^{2}

θ − XY sin

^{2}

θ − Y

^{2}

sin θ cosθ ]

= X

^{2}

+ Y

^{2}

+ 4

\left[ \left( X^{2} - Y^{2} \right) \frac{\sin 2\theta}{2} + XY \cos 2\theta \right]

= X

^{2}

+ Y

^{2}

+ 2( X

^{2}

−Y

^{2}

)sin 2 θ + 4 XY cos 2θ …(ii)

From (i) and (ii) we get

cos2 θ = 0

⇒ θ =

\frac{\pi}{4}

∴ X

^{2}

+ Y

^{2}

+ 2(X

^{2}

− Y

^{2}

) = A X

^{2}

+ BY

^{2}

3 X

^{2}

− Y

^{2}

= A X

^{2}

+ BY

^{2}

⇒ A = 3 ; B = −1

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