Question

Solve sec x+ tan x = cot 4

Answer 1

x = \frac{\pi}{2}-8+2n\pi or x=-\frac{\pi}{2}+2n\pi, n\in\mathbb{Z}

Answer 2

We wish to solve

$$\sec x+\tan x=\cot 4.$$

A step‐by‐step solution is given below.

Step 1. Express in terms of sine and cosine

Recall that

$$\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x}.$$

Thus the equation becomes

$$\frac{1+\sin x}{\cos x}=\cot 4=\frac{\cos 4}{\sin 4}.$$

Step 2. Cross Multiply

Assuming $\cos x\neq 0$ and $\sin 4\neq 0$, cross multiply:

$$(1+\sin x)\sin4=\cos4\,\cos x.$$

Step 3. Use the Tangent Half‐Angle Substitution

A neat way to simplify $\frac{1+\sin x}{\cos x}$ is to use the substitution

$$t=\tan\frac{x}{2}.$$

Recall the identities:

$$\sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2}.$$

Then

$$\frac{1+\sin x}{\cos x}=\frac{1+\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} =\frac{(1+t^2)+2t}{1-t^2} =\frac{(1+t)^2}{1-t^2}.$$

Thus the equation becomes

$$\frac{(1+t)^2}{1-t^2}=\cot 4=\frac{\cos4}{\sin4}.$$

Step 4. Solve for $t$

Write the equation as:

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

Let

$$A=\frac{\cos4}{\sin4} \quad (\text{note } A>0 \text{ if } 4 \text{ is in radians with } \sin4\neq0).$$

Then

$$(1+t)^2=A(1-t^2).$$

Expand the left‐hand side:

$$1+2t+t^2=A-A\,t^2.$$

Rearrange to obtain:

$$(1+A)t^2+2t+(1-A)=0.$$

This is a quadratic in $t$. The solution is given by

$$t=\frac{-2\pm\sqrt{4-4(1+A)(1-A)}}{2(1+A)}.$$

Notice that

$$(1+A)(1-A)=1-A^2,$$

so the discriminant becomes:

$$4-4(1-A^2)=4A^2.$$

Thus,

$$t=\frac{-2\pm 2A}{2(1+A)} =\frac{-1\pm A}{1+A}.$$

This gives two cases:

1. Using the “+” sign: $$t=\frac{-1+A}{1+A}.$$
2. Using the “–” sign: $$t=\frac{-1-A}{1+A}=-1.$$

Since $t=\tan\frac{x}{2}$, we now have:

• Case 1:

  $$\tan\frac{x}{2}=\frac{A-1}{A+1},\quad A=\cot4.$$

  Notice that

  $$\frac{A-1}{A+1}=\frac{\cot4-1}{\cot4+1}.$$

  Write $\cot4=\frac{\cos4}{\sin4}$ so that

  $$\frac{\cot4-1}{\cot4+1} =\frac{\frac{\cos4}{\sin4}-1}{\frac{\cos4}{\sin4}+1} =\frac{\cos4-\sin4}{\cos4+\sin4}.$$

  Now, recalling that

  $$\frac{\cos4-\sin4}{\cos4+\sin4}=\frac{1-\tan4}{1+\tan4}=\tan\left(\frac{\pi}{4}-4\right),$$

  we obtain

  $$\tan\frac{x}{2}=\tan\left(\frac{\pi}{4}-4\right).$$

  Hence,

  $$\frac{x}{2}=\frac{\pi}{4}-4+n\pi,\quad n\in\mathbb{Z},$$

  which implies

  $$x=\frac{\pi}{2}-8+2n\pi,\quad n\in\mathbb{Z}.$$

• Case 2:

  $$\tan\frac{x}{2}=-1.$$

  Therefore,

  $$\frac{x}{2}=-\frac{\pi}{4}+n\pi,\quad n\in\mathbb{Z},$$

  and so

  $$x=-\frac{\pi}{2}+2n\pi,\quad n\in\mathbb{Z}.$$

The solutions of the equation

$$\sec x+\tan x=\cot 4$$

are:

$$x=\frac{\pi}{2}-8+2n\pi\quad\text{or}\quad x=-\frac{\pi}{2}+2n\pi,\quad n\in\mathbb{Z}.$$