P is a point on the segment joining the feet of two vertical poles of heights a and b. The angles of elevation of the tops of the poles from P are 45° each. Then, the square of the distance between the tops of the poles is

• $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2}{2}$

• ${\mathrm{a}}^2 + {\mathrm{b}}^2$

• $2\left({\mathrm{a}}^2 + {\mathrm{b}}^2\right)$

• $4\left({\mathrm{a}}^2 + {\mathrm{b}}^2\right)$

The transformed equation of x² + y² = r² when the axes are rotated through an angle 36° is

• $\sqrt{5}{\mathrm{X}}^2 - 4\mathrm{XY} + {\mathrm{Y}}^2 = {\mathrm{r}}^2$

• ${\mathrm{X}}^2 + 2\mathrm{XY} - \sqrt{5}{\mathrm{Y}}^2 = {\mathrm{r}}^2$

• ${\mathrm{X}}^2 - {\mathrm{Y}}^2 = {\mathrm{r}}^2$

• ${\mathrm{X}}^2 + {\mathrm{Y}}^2 = {\mathrm{r}}^2$

263.

$$\begin{gathered} \mathrm{The} \mathrm{period} \mathrm{of} \left(\tan \left(\mathrm{\theta }\right) - \frac{1}{3}{\tan }^3\left(\mathrm{\theta }\right)\right){\left(\frac{1}{3} - {\tan }^2\left(\mathrm{\theta }\right)\right)}^{ - 1}, \\ \mathrm{where} {\tan }^2\left(\mathrm{\theta }\right) \ne \frac{1}{3} \mathrm{is} \end{gathered}$$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{2\mathrm{\pi }}{3}$

• $\mathrm{\pi }$

• $2\mathrm{\pi }$

$$\begin{gathered} \mathrm{If} {\mathrm{asin}}^2\left(\mathrm{\theta }\right) + {\mathrm{bcos}}^2\left(\mathrm{\theta }\right) = \mathrm{c}, \\ \mathrm{then} {\tan }^2\left(\mathrm{\theta }\right) = ? \end{gathered}$$

• $\frac{\mathrm{b} - \mathrm{c}}{\mathrm{a} - \mathrm{c}}$

• $\frac{\mathrm{c} - \mathrm{b}}{\mathrm{a} - \mathrm{c}}$

• $\frac{\mathrm{a} - \mathrm{c}}{\mathrm{b} - \mathrm{c}}$

• $\frac{\mathrm{a} - \mathrm{c}}{\mathrm{c} - \mathrm{b}}$

If cos(x - y), cos(x), cos(x + y) are three distinct numbers which are in harmoric progression and cos(x) $\ne$ cos(y), then 1 + cos(y) is equal to

• ${\cos }^2\left(\mathrm{x}\right)$

• $- {\cos }^2\left(\mathrm{x}\right)$

• ${\cos }^2\left(\mathrm{x}\right) - 1$

• ${\cos }^2\left(\mathrm{x}\right) - 2$

$\mathrm{The} \mathrm{set} \mathrm{of} \mathrm{solutions} \mathrm{of} \mathrm{the} \mathrm{equation} \left(\sqrt{3} - 1\right)\sin \left(\mathrm{\theta }\right) + \left(\sqrt{3} + 1\right)\cos \left(\mathrm{\theta }\right) = 2 \mathrm{is}$

• $\left\{2\mathrm{n\pi } \pm \frac{\mathrm{\pi }}{4} + \frac{\mathrm{\pi }}{12} \hspace{0.17em}: \mathrm{n} \in \mathrm{Z}\right\}$

• $\left\{2\mathrm{n\pi } \pm \frac{\mathrm{\pi }}{4} - \frac{\mathrm{\pi }}{12} \hspace{0.17em}: \mathrm{n} \in \mathrm{Z}\right\}$

• $\left\{\mathrm{n\pi } + {\left(- 1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4} + \frac{\mathrm{\pi }}{12} \hspace{0.17em}: \mathrm{n} \in \mathrm{Z}\right\}$

• $\left\{\mathrm{n\pi } + {\left(- 1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4} - \frac{\mathrm{\pi }}{12} \hspace{0.17em}: \mathrm{n} \in \mathrm{Z}\right\}$

$$\begin{gathered} \mathrm{If} ∆ = {\mathrm{a}}^2 - {\left(\mathrm{b} - \mathrm{c}\right)}^2, \mathrm{is} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} ∆\mathrm{ABC}, \\ \mathrm{then} \tan \left(\mathrm{A}\right) = ? \end{gathered}$$

• $\frac{1}{16}$

• $\frac{8}{15}$

• $\frac{3}{4}$

• $\frac{4}{3}$

$\mathrm{In} \mathrm{a} ∆∆\mathrm{ABC}, \mathrm{C} = 90°. \mathrm{Then}, \frac{{\mathrm{a}}^2 - {\mathrm{b}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2} = ?$

• sin(A + B)

• sin(A - B)

• cos(A + B)

• cos(A - B)

The sum of angles of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it is 90°. Then, the height of the tower is

• ${\mathrm{a}}^2\mathrm{b}$

• ab²

• $\sqrt{\mathrm{ab}}$

• ab

If 1° = $\mathrm{\alpha }$. radians, then the approximate value of cos(60° 1') is

• $\frac{1}{2} + \frac{\mathrm{\alpha }\sqrt{3}}{120}$

• $\frac{1}{2} - \frac{\mathrm{\alpha }}{120}$

• $\frac{1}{2} - \frac{\mathrm{\alpha }\sqrt{3}}{120}$

• $\frac{1}{2} + \frac{\mathrm{\alpha }}{120}$