Jamia Millia Islamia 2016 Previous Year Questions (PYQs) – Page 2 of 10

$\sec^{2}\theta = 1 + \tan^{2}\theta$ and $\csc^{2}\theta = 1 + \cot^{2}\theta$. So, $\sec^{2}\theta + \csc^{2}\theta = 2 + \tan^{2}\theta + \cot^{2}\theta$. Now, $\tan^{2}\theta + \cot^{2}\theta = \dfrac{\sin^{4}\theta + \cos^{4}\theta}{\sin^{2}\theta \cos^{2}\theta} = \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}.$ Hence, $2\sin^{2}\theta \cos^{2}\theta(\sec^{2}\theta + \csc^{2}\theta) = 2\sin^{2}\theta \cos^{2}\theta \left(2 + \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2\sin^{2}\theta \cos^{2}\theta \left(\dfrac{2\sin^{2}\theta \cos^{2}\theta + 1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2.$

Let $\cos\theta = x \Rightarrow \sec\theta = \dfrac{1}{x}$. So, $x + \dfrac{1}{x} = 3 \Rightarrow x^{2} + \dfrac{1}{x^{2}} = (x + \dfrac{1}{x})^{2} - 2 = 9 - 2 = 7.$ Hence $\cos^{2}\theta + \sec^{2}\theta = 7.$

$\tan1^{\circ} \cdot \tan89^{\circ} = \tan1^{\circ} \cdot \cot1^{\circ} = 1$. Similarly, $\tan2^{\circ} \cdot \tan88^{\circ} = 1$, and so on. But $\tan45^{\circ} = 1$. Hence the entire product = 1.

Given $\cos\theta + \cos^{3}\theta = \sin^{2}\theta$. Now, $\sin^{2}\theta = 1 - \cos^{2}\theta$. So, $1 - \cos^{2}\theta = \cos\theta + \cos^{3}\theta \Rightarrow \cos^{3}\theta + \cos^{2}\theta + \cos\theta - 1 = 0$. Let $\cos\theta = 1$, then $\sin\theta = 0$. Substituting $\sin\theta = 0$ gives $\sin^{6}\theta - 4\sin^{4}\theta + 8\sin^{2}\theta = 0 - 0 + 0 = 0$. But for $\cos\theta = \frac{1}{2}$, $\sin^{2}\theta = \frac{3}{4}$ gives value $4$.

$\cos^{2}\theta + \sec^{2}\theta = \cos^{2}\theta + \dfrac{1}{\cos^{2}\theta}$. Let $x = \cos^{2}\theta$. Then $a = x + \dfrac{1}{x}$. By AM ≥ GM, $x + \dfrac{1}{x} \ge 2$.

$\dfrac{1+i\sqrt{3}}{\sqrt{3}+i} = \dfrac{(1+i\sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} = \dfrac{(\sqrt{3}+3)i + (\sqrt{3}-1)}{4}$. Hence, $\tan\theta = \dfrac{\text{Imag}}{\text{Real}} = \dfrac{\sqrt{3}+3}{\sqrt{3}-1} = \tan\left(\dfrac{\pi}{3}\right)$.

Let $z = x + iy$, then $\bar{z} = x - iy$. Substitute: $(x + iy)^{2} + 2(x - iy) = 0$. $\Rightarrow x^{2} - y^{2} + 2ixy + 2x - 2iy = 0$. Equating real and imaginary parts: Real: $x^{2} - y^{2} + 2x = 0$ Imag: $2xy - 2y = 0 \Rightarrow y( x - 1 ) = 0$. If $y=0$, then $x^{2} + 2x = 0 \Rightarrow x = 0, -2$. If $x=1$, then $1 - y^{2} + 2 = 0 \Rightarrow y^{2} = 3$. Hence, total 4 solutions.

Let $z = x + iy$, then $|z|^{2} = x^{2} + y^{2}$. So, $(x + iy)^{2} + x^{2} + y^{2} = 0$. $\Rightarrow (2x^{2} - y^{2}) + 2ixy = 0 \Rightarrow 2x^{2} - y^{2} = 0, 2xy = 0$. If $x = 0$, then $-y^{2} = 0 \Rightarrow y = 0$. If $y = 0$, then $2x^{2} = 0 \Rightarrow x = 0$. Non-trivial solution: $x = y = 0$. So we take $z = i\sqrt{2}|x|$ form. $z = 2 + 3i$ satisfies the given equation.

We know $\omega^{3} = 1$ and $1 + \omega + \omega^{2} = 0$. Expanding: $(1 + \omega - \omega^{2})(1 - \omega + \omega^{2}) = 1 - \omega^{2} + \omega^{4} - \omega + \omega^{2} - \omega^{3} + \omega^{2} - \omega^{3} + \omega^{4} = 1 + 3 = 3.$

$(x - 1)^{3} = -8 \Rightarrow x - 1 = 2\omega^{k}$ where $k = 0, 1, 2$. So, $x = 1 + 2\omega^{k}$. For $k = 2$, $x = 1 - 2\omega^{2}$.