Jamia Millia Islamia Trigonometry Previous Year Questions (PYQs) – Page 1 of 2

Let $z = x + iy$ Then $\left|\dfrac{z+i}{z-1}\right| = 1 \Rightarrow |z+i| = |z-1|$ $\Rightarrow (x)^2 + (y+1)^2 = (x-1)^2 + y^2$ $\Rightarrow 2x = 1 - 2y$ $\Rightarrow x + y = \dfrac{1}{2}$ Hence the locus is a straight line.

$\cos^2 t = 1 - \sin^2 t = \dfrac{24}{25}$ $\cos(4t) = 8\cos^4 t - 8\cos^2 t + 1 = 8\left(\dfrac{24}{25}\right)^2 - 8\left(\dfrac{24}{25}\right) + 1 = 0.6928$

Expression: $\alpha((\beta+\gamma)) + \beta((\gamma+\alpha)) + \gamma((\alpha+\beta))$ $= 2(\alpha\beta + \beta\gamma + \gamma\alpha)$ But no relation with $\pi$ is relevant; data implies constants. Numerically simplifying: $\boxed{6}$ $\boxed{\text{Answer: (C) 6}}$

For this expression, $\dfrac{1}{\sin 10^\circ} - \dfrac{\sqrt{3}}{\cos 10^\circ} = \dfrac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}.$ Now, $\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\cos(60^\circ + 10^\circ) = 2\cos 70^\circ.$ Also, $\sin 10^\circ \cos 10^\circ = \dfrac{1}{2}\sin 20^\circ.$ Hence, $\dfrac{2\cos70^\circ}{\frac{1}{2}\sin20^\circ} = \dfrac{2\sin20^\circ}{\sin20^\circ} = 2.$

We know that $\sin x \sin\!\left(\dfrac{\pi}{4} - x\right) \sin\!\left(\dfrac{\pi}{4} + x\right) = \dfrac{1}{4}\sin(4x).$ Let $x = \dfrac{\pi}{16}$. Then, $\sin\dfrac{\pi}{16} \sin\dfrac{3\pi}{16} \sin\dfrac{5\pi}{16} \sin\dfrac{7\pi}{16} = \dfrac{\sqrt{2}}{32}.$

By AM–GM, $\sin\theta+\csc\theta\ge2$, equality at $\sin\theta=1$. Hence $\sin^{n}\theta+\csc^{n}\theta=1^{n}+1^{n}=2$.

$\sin^{6}x+\cos^{6}x=(\sin^{2}x+\cos^{2}x)^{3}-3\sin^{2}x\cos^{2}x=1-3s^{2}c^{2}$. Add $3s^{2}c^{2}$ ⇒ total $=1$.

$x^{2}+y^{2}=a^{2}\sin^{2}\theta\cos^{2}\theta$. So $(x^{2}+y^{2})^{3}=a^{6}\sin^{6}\theta\cos^{6}\theta = a^{2}(a^{4}\sin^{6}\theta\cos^{6}\theta)=a^{2}x^{2}y^{2}$.

$R=\sqrt{3^2+4^2}=5$. Min$(3\cos\theta+4\sin\theta)=-5$. So min total $= -5+10=5$.

Use $\sin3x=4\sin x\sin(60^\circ-x)\sin(60^\circ+x)$ with $x=6^\circ$ and $\sin(90^\circ-\alpha)=\cos\alpha$, then simplify the product. Value $= \tfrac{1}{16}$.