General value of theta satisfying the equati | vedclass

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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

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