Question

If $sin(\frac{\pi}{4}cot\theta)=cos(\frac{\pi}{4}tan\theta)$, then $\theta=$

• A. $n\pi + \frac{\pi}{4}$
• B. $2n\pi \pm \frac{\pi}{4}$
• C. $n\pi - \frac{\pi}{4}$
• D. $2n\pi \pm \frac{\pi}{6}$

Answer 1

Answer: (A)

$n\pi + \frac{\pi}{4}$

Answer 2

The equation $sin(\frac{\pi}{4}cot\theta)=cos(\frac{\pi}{4}tan\theta)$ can be rewritten using the identity $cos(x) = sin(\frac{\pi}{2} - x)$:

$sin(\frac{\pi}{4}cot\theta) = sin(\frac{\pi}{2} - \frac{\pi}{4}tan\theta)$

The general solution for $sin(A) = sin(B)$ is $A = n\pi + (-1)^n B$. Let $A = \frac{\pi}{4}cot\theta$ and $B = \frac{\pi}{2} - \frac{\pi}{4}tan\theta$.

Case 1: $n$ is even ($n=2k$). $\frac{\pi}{4}cot\theta = 2k\pi + (\frac{\pi}{2} - \frac{\pi}{4}tan\theta)$ Dividing by $\frac{\pi}{4}$: $cot\theta = 8k + 2 - tan\theta$ $cot\theta + tan\theta = 8k + 2$ $\frac{cos\theta}{sin\theta} + \frac{sin\theta}{cos\theta} = 8k + 2$ $\frac{cos^2\theta + sin^2\theta}{sin\theta cos\theta} = 8k + 2$ $\frac{1}{sin\theta cos\theta} = 8k + 2$ $\frac{2}{sin(2\theta)} = 8k + 2$ $sin(2\theta) = \frac{2}{8k+2} = \frac{1}{4k+1}$

For $k=0$, $sin(2\theta) = 1$. The general solution is $2\theta = 2m\pi + \frac{\pi}{2}$. $\theta = m\pi + \frac{\pi}{4}$. This matches option (A).

Case 2: $n$ is odd ($n=2k+1$). $\frac{\pi}{4}cot\theta = (2k+1)\pi - (\frac{\pi}{2} - \frac{\pi}{4}tan\theta)$ Dividing by $\frac{\pi}{4}$: $cot\theta = 4(2k+1) - 2 + tan\theta$ $cot\theta = 8k + 4 - 2 + tan\theta$ $cot\theta - tan\theta = 8k + 2$ $\frac{cos^2\theta - sin^2\theta}{sin\theta cos\theta} = 8k + 2$ $\frac{cos(2\theta)}{sin\theta cos\theta} = 8k + 2$ $\frac{2cos(2\theta)}{sin(2\theta)} = 2(8k+2)$ $2cot(2\theta) = 16k+4$ $cot(2\theta) = 8k+2$ This case does not yield any of the given options.

Checking option (A): If $\theta = n\pi + \frac{\pi}{4}$, then $tan\theta = 1$ and $cot\theta = 1$. LHS = $sin(\frac{\pi}{4} \times 1) = sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. RHS = $cos(\frac{\pi}{4} \times 1) = cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. LHS = RHS, so option (A) is correct.