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

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?

• What did you get for sin($t$) and tan($t$)? Commented Feb 11, 2019 at 4:37

Hint:

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

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

• More succinct version of mine. Faster to enter, too. Commented Feb 11, 2019 at 4:40
• Why I prefer writing hints to full answers. Also makes the OP think. Commented Feb 11, 2019 at 4:41
• Thanks, Rhys. I am reviewing trigonometry and I think I forgot almost everything. Commented Feb 11, 2019 at 4:46

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

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 $$\dfrac1{2a}$$.

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: $$\lambda +\dfrac{1}{\lambda}=2\sec t =2a+\dfrac{1}{2a}\implies \lambda=2a \ \text{or} \dfrac{1}{2a} \text{.}$$

• This is my way(+1) Commented Feb 11, 2019 at 5:36

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

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

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

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

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.