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Question

Write the function $$\tan^{-1} \left (\dfrac {\cos x - \sin x}{\cos x + \sin x}\right ), x < \pi$$ in the simplest form


Solution

$$tan^{-1}(\dfrac{cosx-sinx}{cosx+sinx})$$

$$=tan^{-1}(\dfrac{1-tanx}{1+tanx})$$

$$=tan^{-1}(\dfrac{tan45^{0}-tanx}{1+tanx tan45^{0}})$$

$$=tan^{-1}(tan(45^{0}-x))$$
Now
$$tan(45^{0}-x)>0$$ for
$$90^{0}>45^0-x>0$$
Or $$45^{0}>x>-45^{0}$$.
Since $$x<180^{0}$$ hence for $$45^{0}>x>-45^{0}$$
$$tan^{-1}(tan(45^{0}-x))=45^{0}-x$$

Mathematics

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