If u=sec2theta+tan2theta, find (u-1)/(u+1)?

Dec 13, 2016

Explanation:

We are given $u = \sec 2 \theta + \tan 2 \theta$....(1)

and hence using componendo and dividendo

$\frac{u - 1}{u + 1} = \frac{\sec 2 \theta + \tan 2 \theta - 1}{\sec 2 \theta + \tan 2 \theta + 1}$

multiplying numerator and denominator on RHS by $\cos 2 \theta$, we get

$\frac{u - 1}{u + 1} = \frac{1 + \sin 2 \theta - \cos 2 \theta}{1 + \sin 2 \theta + \cos 2 \theta}$

= $\frac{1 + 2 \sin \theta \cos \theta - {\cos}^{2} \theta + {\sin}^{2} \theta}{1 + 2 \sin \theta \cos \theta + {\cos}^{2} \theta - {\sin}^{2} \theta}$

$= \frac{{\sin}^{2} \theta + 2 \sin \theta \cos \theta + {\sin}^{2} \theta}{{\cos}^{2} \theta + 2 \sin \theta \cos \theta + {\cos}^{2} \theta}$

$= \frac{2 {\sin}^{2} \theta + 2 \sin \theta \cos \theta}{2 {\cos}^{2} \theta + 2 \sin \theta \cos \theta}$

=(2sintheta(sintheta+costheta))/(2costheta(costheta+sintheta)

= $\frac{2 \sin \theta}{2 \cos \theta}$

= $\tan \theta$

= $t$

Jun 19, 2018

$\tan \theta$.

Explanation:

Prerequisite : Componendo Dividendo (CD) :

$C D : \frac{a}{b} = \frac{c}{d} \iff \frac{a - b}{a + b} = \frac{c - d}{c + d}$.

We have, $u = \sec 2 \theta + \tan 2 \theta$,

$= \frac{1}{\cos 2 \theta} + \frac{\sin 2 \theta}{\cos 2 \theta}$,

$= \frac{1 + \sin 2 \theta}{\cos 2 \theta}$,

$= \frac{\left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + 2 \cos \theta \sin \theta}{{\cos}^{2} \theta - {\sin}^{2} \theta}$,

$= {\left(\cos \theta + \sin \theta\right)}^{\cancel{2}} / \left\{\cancel{\left(\cos \theta + \sin \theta\right)} \left(\cos \theta - \sin \theta\right)\right\}$,

$\Rightarrow \frac{u}{1} = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta}$.

$\text{By CD, then, } \frac{u - 1}{u + 1}$,

$= \frac{\left(\cos \theta + \sin \theta\right) - \left(\cos \theta - \sin \theta\right)}{\left(\cos \theta + \sin \theta\right) + \left(\cos \theta - \sin \theta\right)}$,

$= \frac{2 \sin \theta}{2 \cos \theta}$,

$\Rightarrow \frac{u - 1}{u + 1} = \tan \theta$,

as Respected Shwetank Mauria has readily derived!

Jun 19, 2018

$\tan \theta .$ Kindly go through Explanation.

Explanation:

Here is a Third Method to solve the Problem, if we use,

$\cos 2 \theta = \frac{1 - {t}^{2}}{1 + {t}^{2}} , \mathmr{and} , \tan 2 \theta = \frac{2 t}{1 - {t}^{2}} , t = \tan \theta$.

Hence, $u = \sec 2 \theta + \tan 2 \theta = \frac{1}{\cos 2 \theta} + \tan 2 \theta$,

$= \frac{1 + {t}^{2}}{1 - {t}^{2}} + \frac{2 t}{1 - {t}^{2}}$,

$= \frac{1 + 2 t + {t}^{2}}{1 - {t}^{2}}$,

$= {\left(1 + t\right)}^{2} / \left\{\left(1 + t\right) \left(1 - t\right)\right\}$.

$\Rightarrow \frac{u}{1} = \frac{1 + t}{1 - t}$.

$\Rightarrow \frac{u - 1}{u + 1} = \frac{2 t}{2} = t = \tan \theta$, as Before!