MCQ
The value of $\tan ( - 945^\circ )$ is
  • $-1$
  • B
    $-2$
  • C
    $-3$
  • D
    $-4$

Answer

Correct option: A.
$-1$
a
(a) $\tan ( - 945^\circ ) = \tan [ - (945^\circ )]$

$ = - \tan [(2 \times 360^\circ + 225^\circ )]$

$ = - \tan [225^\circ ] = - \tan 45^\circ = - 1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $\mathrm{A}\,(\sec \theta, 2 \tan \theta)$ and $\mathrm{B}\,(\sec \phi, 2 \tan \phi)$, where $\theta+\phi=\pi / 2$, be two points on the hyperbola $2 \mathrm{x}^{2}-\mathrm{y}^{2}=2$. If $(\alpha, \beta)$ is the point of the intersection of the normals to the hyperbola at $\mathrm{A}$ and $\mathrm{B}$, then $(2 \beta)^{2}$ is equal to ..... .
The sum of the series $\frac{1}{{3 \times 7}} + \frac{1}{{7 \times 11}} + \frac{1}{{11 \times 25}} + ....$ is
$(b \times c) \times (c \times a) = $
If $y(x)$ satisfies the differential equation $y' + y = 2(sinx + cosx)$ and $y(0) = 1$, then
If ${n^{th}}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$, then the ratio of their ${12^{th}}$ terms will be
Let $d \in R$, and  $A = \left[ {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {\sin \,\theta } \right) - 2}\\ 1&{\left( {\sin \,\theta } \right) + 2}&d\\ 5&{\left( {2\sin \,\theta } \right) - d}&{\left( { - \sin \,\theta } \right) + 2 + 2d} \end{array}} \right]$, $\theta  \in \left[ {0,2\pi } \right]$. If the minimum value of det $(A)$ is $8$, then a value of $d$ is
Let $g$ $(x)$ be an antiderivative for $f$ $(x)$. Then $ln ( 1+ (g(x))^2)$ is an antiderivative for
The maximum value of $e^{(2 + \sqrt 3 \cos x + \sin x)}$ is
Let $P$ be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let the line passing through $P$ and parallel to $y$-axis meet the circle $x^2+y^2=9$ at point $Q$ such that $P$ and $Q$ are on the same side of the $x$-axis. Then, the eccentricity of the locus of the point $R$ on $P Q$ such that $P R: R Q=4: 3$ as $P$ moves on the ellipse, is :
Given that $\pi < \alpha < \frac{{3\pi }}{2},$ then the expression $\sqrt {(4{{\sin }^4}\alpha + {{\sin }^2}2\alpha )} + 4{\cos ^2}\left( {\frac{\pi }{4} - \frac{\alpha }{2}} \right)$ is equal to