Question

If \(x\) is a real number such that \(tanx+cotx=2\),then \(x=\)?

• A. \( (n+\dfrac{1}{4})π,n∈Z\)
• B. \((n+1)π,n∈Z\)
• C. \( (n+\dfrac{1}{2})π,n∈Z\)
• D. \(nπ,n∈Z\)
• E. \( \dfrac{2}{3}nπ,n∈Z\)

Answer 1

The Correct Option is A

Given:
Here , x is the real number 

\(tanx+cotx=2\)

⇒\(tanx\)\(+\dfrac{1}{tanx}=2\)

⇒\(\dfrac{(tanx)^{2}+1}{tan(x)}=2\)

⇒\((tanx)^{2}-2tanx+1=0\)

⇒\((tanx-1)^{2}=0\)

⇒\(tanx=±1\)

⇒\(x=\tan^{-1} (1)\)

\(x=\dfrac{\pi}{4}\\\text  or \\\text = 3\dfrac{3\pi}{4}\)

Hence ,the correct option is \((n+\dfrac{1}{4})π,n∈Z\)  

Answer 2

We are given that \( \tan(x) + \cot(x) = 2 \). We can rewrite this equation using the definitions of tangent and cotangent:

\[ \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = 2 \]

To solve this, we find a common denominator:

\[ \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} = 2 \]

Since \( \sin^2(x) + \cos^2(x) = 1 \), the equation simplifies to:

\[ \frac{1}{\sin(x)\cos(x)} = 2 \] \[ 1 = 2\sin(x)\cos(x) \]

Using the double angle identity, \( 2\sin(x)\cos(x) = \sin(2x) \), we have:

\[ \sin(2x) = \frac{1}{2} \]

This means:

\[ 2x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad 2x = \frac{5\pi}{6} + 2n\pi, \quad \text{where } n \text{ is an integer.} \]

Solving for \( x \):

\[ x = \frac{\pi}{12} + n\pi \quad \text{or} \quad x = \frac{5\pi}{12} + n\pi \]

Now let's examine the given options:

• \((n + \frac{1}{4})\pi\): This matches \( x = \frac{\pi}{12} + n\pi \) if \( n \) is an integer.
• \((n + 1)\pi\): This is not consistent with our solution.
• \((n + \frac{1}{2})\pi\): This is not consistent with our solution.

Therefore, the solution is \( x = (n + \frac{1}{4})\pi \), where \( n \) is an integer.