>
Exams
>
Mathematics
>
Limits
>
by considering 1 0 0175 the approximate value of c
Question:
By considering \( 1' = 0.0175 \), the approximate value of \( \cot 45^\circ 2' \) is:
Show Hint
For small angles, use the approximation \( \cot(45^\circ + \theta) \approx \frac{1 - \theta}{1 + \theta} \) to estimate values efficiently.
AP EAMCET - 2024
AP EAMCET
Updated On:
Mar 19, 2025
\( 1.07 \)
\( 0.965 \)
\( 1.035 \)
\( 0.93 \)
Show Solution
Verified By Collegedunia
The Correct Option is
D
Solution and Explanation
Step 1: Using small-angle approximation
We use: \[ \cot(45^\circ + \theta) \approx \frac{1 - \theta}{1 + \theta}. \] Substituting \( \theta = 2' = 2 \times 0.0175 \), we compute: \[ \cot 45^\circ 2' \approx \frac{1 - 0.035}{1 + 0.035} = \frac{0.965}{1.035} \approx 0.93. \]
Download Solution in PDF
Was this answer helpful?
0
0
Top AP EAMCET Mathematics Questions
If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
AP EAMCET - 2019
Mathematics
Definite Integral
View Solution
If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
AP EAMCET - 2019
Mathematics
Transpose of a Matrix
View Solution
If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
AP EAMCET - 2019
Mathematics
Maxima and Minima
View Solution
If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
AP EAMCET - 2019
Mathematics
Trigonometric Equations
View Solution
If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is a focal chord, then one possible values of $\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}$ is
AP EAMCET - 2019
Mathematics
Ellipse
View Solution
View More Questions
Top AP EAMCET Limits Questions
If $\alpha =\displaystyle\lim_{x\to0} \frac{x.2^{x}-x}{1-\cos x} $ and $ \beta =\displaystyle\lim_{x\to0} \frac{x.2^{x}-x}{\sqrt{1+x^{2} } - \sqrt{1-x^{2}} } , $ then
AP EAMCET - 2019
Mathematics
Limits
View Solution
$\lim_{x \to \frac{\pi}{2}} \frac{1+ \cos2x }{\cot3x \left(3^{\sin2x} - 1\right)} = $
AP EAMCET - 2018
Mathematics
Limits
View Solution
Let \( f(x) = \begin{cases
1 + \frac{2x}{a}, & 0 \le x \le 1
ax, & 1<x \le 2 \end{cases} \). If \( \lim_{x \to 1} f(x) \) exists, then the sum of the cubes of the possible values of \( a \) is: }
AP EAMCET - 2024
Mathematics
Limits
View Solution
Let \( [P] \) denote the greatest integer \( \leq P \). If \( 0 \leq a \leq 2 \), then the number of integral values of \( a \) such that \( \lim_{x \to a} [x^2] - [x]^2 \) does not exist is:
AP EAMCET - 2024
Mathematics
Limits
View Solution
If \( f(x) = \begin{cases
\frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{\sqrt{a + x} - \sqrt{a - x}}, & x \ne 0
K, & x = 0 \end{cases} \) is continuous at \( x = 0 \), then \( K = \)}
AP EAMCET - 2024
Mathematics
Limits
View Solution
View More Questions
Top AP EAMCET Questions
If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
AP EAMCET - 2019
Definite Integral
View Solution
If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
AP EAMCET - 2019
Transpose of a Matrix
View Solution
If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$
AP EAMCET - 2019
Electrical Instruments
View Solution
If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
AP EAMCET - 2019
Maxima and Minima
View Solution
If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
AP EAMCET - 2019
Trigonometric Equations
View Solution
View More Questions