Value of θ lying between θ = 0 and θ = π /2 and satisfying equation : left| {,begin{array}

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

normal

The value of $\theta$ lying between $\theta = 0$ and $\theta = \pi /2$ and satisfying the equation : $\left| {\,\begin{array}{*{20}{c}} {1\,\, + \,\,{{\sin }^2}\,\theta }&{{{\cos }^2}\,\theta }&{4\,\sin \,4\,\theta }\\ {{{\sin }^2}\,\theta }&{1\,\, + \,\,{{\cos }^2}\,\theta }&{4\,\sin \,4\,\theta }\\ {{{\sin }^2}\,\theta }&{{{\cos }^2}\,\theta }&{1\,\, + \,\,4\,\sin \,4\,\theta } \end{array}\,} \right|$ $= 0$ are :

A

$\frac{{7\,\pi }}{{24}}$

B

$\frac{{5\,\pi }}{{24}}$

C

$\frac{{11\,\pi }}{{24}}$

D

both $(A)$ and $(C)$

Solution

Solution is Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut elit tellus, luctus nec ullamcorper mattis, pulvinar dapibus leo.

Standard 12

Mathematics

Share

Similar Questions

Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

easy

Let $\beta$ be a real number. Consider the matrix

$A=\left(\begin{array}{ccc}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{array}\right)$

If $A^7-(\beta-1) A^6-\beta A^5$ is a singular matrix, then the value of $9 \beta$ is

hard

(IIT-2022)

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

hard

(JEE MAIN-2019)

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ – 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $…..$

hard

(JEE MAIN-2021)

If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right|$ $= 0$, then the triangle is

normal