Question
${e^{{e^{ - i\theta }}}}$का कोणांक होगा
$z = {e^{\cos \theta }}[\cos (\sin \theta ) - i\sin (\sin \theta )]$
$z = {e^{\cos \theta }}\cos (\sin \theta ) - i{e^{\cos \theta }}\sin (\sin \theta )$
$amp(z) = {\tan ^{ - 1}}\left[ { - \frac{{{e^{\cos \theta }}\sin (\sin \theta )}}{{{e^{\cos \theta }}\cos (\sin \theta )}}} \right]$
$ = {\tan ^{ - 1}}[\tan ( - \sin \theta )] = - \sin \theta $.
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$f(x)=\left\{\begin{array}{ll}a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4\end{array}\right.$ कुछ $a, b, c \in R$ के लिए
सतत् है और $f^{\prime}(0)+f^{\prime}(2)=e$, तो $a$ का मान है