$$ \text { The expression } \frac{1-\tan ^2\left(\frac{\pi}{4}-A\right)}{1+\tan ^2\left(\frac{\pi}{4}-A\right)} \text { equals } $$
$\sin A$
$\sin 2 A$
$\cos A$
$\cos 2 A$
The behaviour of the function $f(x)=\sin \left(2 x+\frac{\pi}{4}\right)$ on $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$ is:
Strictly increasing on $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$
Strictly increasing on $\left(\frac{\pi}{8}, \frac{3 \pi}{8}\right)$
Strictly decreasing on $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$
${\text { Strictly decreasing on }}\left(\frac{\pi}{8}, \frac{5 \pi}{8}\right)$
$$ \text { The particular solution of the equation } \sin \left(\frac{d y}{d x}\right)=a \text {, where } a \in \mathbb{R} \text { and } y=2 \text { when } x=0 \text { is } $$
$$ \sin \left(\frac{y-2}{x}\right)=a $$
$$ \sin \left(\frac{x-2}{y}\right)=a $$
$$ \sin \left(\frac{y+2}{x}\right)=a $$
$$ \sin \left(\frac{x}{y-2}\right)=a $$
$$ \begin{aligned} &\text { Find the co-ordinates of the orthocentre of the triangle formed by the lines }\\ &\begin{aligned} & L_1: y-x=2 \\ & L_2: y+2 x=8 \\ & L_3: 3 y-x=18 \end{aligned} \end{aligned} $$
$$ \left(\frac{10}{7}, \frac{40}{7}\right) $$
(6, 8)
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