General value of theta satisfying the equati | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) Using $\sec 2	heta = \frac{1}{{\cos 2	heta }} = \frac{{1 + {{	an }^2}	heta }}{{1 - {{	an }^2}	heta }}$,

we can write the given equation a

Vedclass · Accepted Answer

$m\pi ,n\pi \pm \frac{\pi }{3}$

Vedclass · Answer

(b) Using $\sec 2	heta = \frac{1}{{\cos 2	heta }} = \frac{{1 + {{	an }^2}	heta }}{{1 - {{	an }^2}	heta }}$,

we can write the given equation as ${	an ^2}	heta + \frac{{1 + {{	an }^2}	heta }}{{1 - {{	an }^2}	heta }} = 1$.

$ \Rightarrow $ ${	an ^2}	heta (1 - {	an ^2}	heta ) + 1 + {	an ^2}	heta = 1 - {	an ^2}	heta $

$ \Rightarrow $ $3{	an ^2}	heta - {	an ^4}	heta = 0 \Rightarrow {	an ^2}	heta (3 - {	an ^2}	heta ) = 0$

$ \Rightarrow $ $	an 	heta = 0$ or $	an 	heta = \pm \sqrt 3 $

Now $	an 	heta = 0 \Rightarrow 	heta = m\pi $, where $m$ is an integer and

$	an 	heta = \pm \sqrt 3 = 	an ( \pm \pi /3) \Rightarrow 	heta = n\pi \pm \frac{\pi }{3}$, where $n$ is an integer.

Thus $	heta = m\pi ,\,n\pi \pm \frac{\pi }{3}$, where $m$ and $n$ are integers.

General value of $\theta $ satisfying the equation ${\tan ^2}\theta + \sec 2\theta - = 1$ is

Similar Questions

The expression $(1 + \tan x + {\tan ^2}x)$ $(1 - \cot x + {\cot ^2}x)$ has the positive values for $x$, given by

The solution set of $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$ in the interval $[0,\,\,2\pi ]$ is

The general solution of the equation $(\sqrt 3 - 1)\sin \theta + (\sqrt 3 + 1)\cos \theta = 2$ is

If ${\sin ^2}\theta - 2\cos \theta + \frac{1}{4} = 0,$ then the general value of $\theta $ is

If $\cos \theta + \sec \theta = \frac{5}{2}$, then the general value of $\theta $ is