What is the value of e^{pi/2}-e^{arctan(1+(0)^2)} ?

The value of e^{pi/2}-e^{arctan(1+(0)^2)} is e^{pi/2}-e^{pi/4}

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e^{pi/2}-e^{arctan(1+(0)^2)}

e^{\frac{π}{2}} - e^{a r c t a n (1 + (0)^{2})}

e^{\frac{π}{2}} - e^{\frac{π}{4}}

Decimal Notation

149.95

Solution steps

e^{\frac{π}{2}} - e^{a r c t a n (1 + (0)^{2})}

= e^{\frac{π}{2}} - e^{a r c t a n (1 + 0^{2})}

Use the following trivial identity:

arctan (1 + 0^{2}) = \frac{π}{4}

arctan (1 + 0^{2})

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= e^{\frac{π}{2}} - e^{\frac{π}{4}}

\frac{sin ( 2 0 ^{\circ} )}{cos ( 2 0 ^{\circ} )}

cos^{2} (66. 5^{\circ})

4 cos (2 \cdot 0)

\frac{500}{sin ( 1 2 ^{\circ} )}

arccos (0.66)

What is the value of e^{pi/2}-e^{arctan(1+(0)^2)} ?
The value of e^{pi/2}-e^{arctan(1+(0)^2)} is e^{pi/2}-e^{pi/4}