If $\sec(t) = a + 1/(4a)$, prove that $\sec(t) + \tan(t) = 2a$ or $1/(2a)$

Question

I tried to convert $\sec(t)$ into $1/\cos(t)$ and figure out the value for $\sin(t)$. Then I tried to figure out $\tan(t)$ with the values I got but I don't seem to be able to get $2a$ or $1/2a$.

What would be the best approach to solve this question?

Answer 1

Hint:

$$(a+\frac{1}{4a})^2-1=(a-\frac{1}{4a})^2$$

Recall that $\sec^2\theta -1 =\tan^2\theta$

Answer 2

$\sec(t) = a+\dfrac1{4a}$.

Since

$\begin{array}\\ \tan^2(t)+1 &=\dfrac{\sin^2(t)}{\cos^2(t)}+1\\ &=\dfrac{\sin^2(t)+\cos^2(t)}{\cos^2(t)}\\ &=\dfrac{1}{\cos^2(t)}\\ &=\sec^2(t)\\ \end{array} $

so

$\begin{array}\\ \tan^2(t) &=\sec^2(t)-1\\ &=(a+\dfrac1{4a})^2-1\\ &=a^2+\dfrac12+\dfrac1{16a^2}-1\\ &=a^2-\dfrac12+\dfrac1{16a^2}\\ &=(a-\dfrac1{4a})^2\\ \text{so}\\ \tan(t) &=\pm(a-\dfrac1{4a})\\ \end{array} $

Therefore $\tan(t)+\sec(t) =a+\dfrac1{4a}\pm(a-\dfrac1{4a}) =2a $ or $\dfrac1{2a}$.

Answer 3

Let $\sec t+\tan t=\lambda$. From the simple Algebraic identity $[x^2-y^2=(x-y)(x+y)]$ and the Pythagorean identity involving secant and tangent functions: $$\sec^2t-\tan^2t=(\sec t+\tan t)(\sec t-\tan t)=1$$ You can rewrite $\sec t-\tan t=\dfrac{1}{\sec t+\tan t}=\dfrac{1}{\lambda}$. Summing up the expressions we get: $$\lambda +\dfrac{1}{\lambda}=2\sec t =2a+\dfrac{1}{2a}\implies \lambda=2a \ \text{or} \dfrac{1}{2a} \text{.}$$

Answer 4

$$(2a)^2-2(2a)\sec t+1=0$$

$$2a=\dfrac{2\sec t\pm\sqrt{(2\sec t)^2-4}}2=\sec t\pm\tan t$$

If $\sec t-\tan t=2a,\sec t+\tan t=\dfrac1{\sec t-\tan t}=?$

Else $\sec t+\tan t=2a$

Answer 5

You asked for the best approach. See other answers for that. I just wanted to show that it could be done the way you had tried, using only very basic identities:

$$\frac 1 {\cos t} = \sec t = a + \frac 1 {4a} = {4a^2+1 \over 4a}$$

so

$$\cos t = {4a \over {4a^2+1}}$$

so

$$\cos^2 t = {(4a)^2 \over {(4a^2+1)^2}}$$

so

$$\sin^2 t = 1 - \cos^2 t = {(4a^2+1)^2-(4a)^2 \over (4a^2+1)^2} = {(4a^2-1)^2 \over (4a^2+1)^2}$$

so

$$\sin t = \pm \frac{4a^2-1}{4a^2+1}$$

so $$\tan t = {\frac {\sin t} {\cos t}}= \pm {4a^2-1\over 4a} = \pm (a - {{1} \over{4a}}),$$

from which the expression given for $\sec t + \tan t$ easily follows.