Question

cot(pi/4+x)cot(pi/4-theta)=

Answer 1

Here is your solution

Answer 2

Answer:

$\red { cot \left( \frac{\pi}{4} + \theta\right) cot \left( \frac{\pi}{4} - \theta\right)}\green {=1}$

Step-by-step explanation:

$LHS = cot \left( \frac{\pi}{4} + \theta\right) cot \left( \frac{\pi}{4} - \theta\right)$

$= cot \left( \frac{\pi}{4} + \theta\right) cot [\frac{\pi}{2} -\left( \frac{\pi}{4} + \theta\right)]$

$= cot \left( \frac{\pi}{4} + \theta\right) tan \left( \frac{\pi}{4} + \theta\right)$

$= 1$

$\boxed { \pink { cotA \times tanA = 1 }}$

$= RHS$

Therefore.,

$\red { cot \left( \frac{\pi}{4} + \theta\right) cot \left( \frac{\pi}{4} - \theta\right)}\green {=1}$

•••♪