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Question Number 9800 by richard last updated on 05/Jan/17

is there a value for tg^2 (π/4)  ?

$$\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{value}\:\mathrm{for}\:\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}\:\:? \\ $$

Commented by FilupSmith last updated on 05/Jan/17

tg^2 (π/4)=tg^2 (π/4)     unless t and g have values, you can  no further evaluate

$${tg}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}={tg}^{\mathrm{2}} \frac{\pi}{\mathrm{4}} \\ $$$$\: \\ $$$$\mathrm{unless}\:{t}\:\mathrm{and}\:{g}\:\mathrm{have}\:\mathrm{values},\:\mathrm{you}\:\mathrm{can} \\ $$$$\mathrm{no}\:\mathrm{further}\:\mathrm{evaluate} \\ $$

Commented by geovane10math last updated on 06/Jan/17

tg = tan

$$\mathrm{tg}\:=\:\mathrm{tan}\: \\ $$

Answered by sandy_suhendra last updated on 05/Jan/17

do you mean tg = tan?  tan^2  (π/4)=(tan (π/4))^2  = 1^2  = 1

$$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{tg}\:=\:\mathrm{tan}? \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{4}}=\left(\mathrm{tan}\:\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} \:=\:\mathrm{1}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$

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