Let M and m respectively be the maximum and minimum values of the function
f(x) = tan$-$1 (sin x + cos x) in $\left[ {0,{\pi \over 2}} \right]$, then the value of tan(M $-$ m) is equal to :
Solution
Let g(x) = sin x + cos x = $\sqrt 2$ sin$\left( {x + {\pi \over 4}} \right)$<br><br>g(x)$\in$ $\left[ {1,\sqrt 2 } \right]$ for x$\in$ [0, $\pi$/2]<br><br>f(x) = tan<sup>$-$1</sup> (sin x + cos x) $\in$ $\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$<br><br>tan$$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2 $$
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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