What is the general solution for e^{pi/2}-e^{arctan(1+x^2)}=0 ?

The general solution for e^{pi/2}-e^{arctan(1+x^2)}=0 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Add

e^{a r c t a n (1 + x^{2})}

to both sides

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

Simplify

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

Apply exponent rules

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

a^{f (x)} = a^{g (x)}

, then

f (x) = g (x)

\frac{π}{2} = arctan (1 + x^{2})

\frac{π}{2} = arctan (1 + x^{2})

Switch sides

arctan (1 + x^{2}) = \frac{π}{2}

- \frac{π}{2} < arctan (x) < \frac{π}{2}

No Solution for x \in R

3 cot (θ) - 7 = 0

cos (x) = \frac{1249}{1250}

sec (θ) = \frac{9}{7}

4 cos (x - π) = 4

cot (x) (tan (x) - \frac{3}{3}) = 0

What is the general solution for e^{pi/2}-e^{arctan(1+x^2)}=0 ?
The general solution for e^{pi/2}-e^{arctan(1+x^2)}=0 is No Solution for x\in\mathbb{R}