To find: Value of tan-1\(\left(\cfrac{cos\,\text x+sin\,\text x}{cos\,\text x-sin\,\text x}\right)\)
tan-1(cos x + sin x)/(cos x - sin x)
Formula used: (i) tan(A + B) = \(\cfrac{tanA+tanB}{1-tanAtanB}\)
We have, tan-1\(\left(\cfrac{cos\,\text x+sin\,\text x}{cos\,\text x-sin\,\text x}\right)\)
Dividing numerator and denominator by cos x
Now, we can see that tan-1\(\left(\cfrac{cos\,\text x+sin\,\text x}{cos\,\text x-sin\,\text x}\right)\) = \(\cfrac{n}4+\text x\)
Now differentiating,