Question
$\int_0^{\pi /2} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}} \,dx = $
अब $I = \int_0^{\pi /2} {\frac{{\cos \left( {\frac{\pi }{2} - x} \right) - \sin \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}\,dx} $
$= \int_0^{\pi /2} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}\,\,dx} $.....$(ii)$
जोड़़ने पर, $2I = 0 \Rightarrow I = 0$.
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$(A)$ $1-\sqrt{\frac{3}{2}}$ $(B)$ $1+\sqrt{\frac{3}{2}}$ $(C)$ $1-\sqrt{\frac{2}{3}}$ $(D)$ $1+\sqrt{\frac{2}{3}}$
$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$
तो $n$ का मान है