Whakaoti i te 5+6=frac{1-sin^{2}45^{circ}}{1+sin^245^circ}+tan^245^circ

$$5 + 6 = \frac { 1 - \sin ^ { 2 } 45 ^ { \circ } } { 1 + \sin ^ { 2 } 45 ^ { \circ } } + \tan ^ { 2 } 45 ^ { \circ }$$

Manatoko

Ngā Upane Otinga

Tāpirihia te 5 ki te 6, ka 11.

$$11=\frac{1-\left(\sin(45)\right)^{2}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Tīkina te uara \sin(45) mai i te ripanga uara pākoki.

$$11=\frac{1-\left(\frac{\sqrt{2}}{2}\right)^{2}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Kia whakarewa i te \frac{\sqrt{2}}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$11=\frac{1-\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$11=\frac{1-\frac{2}{2^{2}}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$11=\frac{1-\frac{2}{4}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$11=\frac{1-\frac{1}{2}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Tangohia te \frac{1}{2} i te 1, ka \frac{1}{2}.

$$11=\frac{\frac{1}{2}}{1+\left(\sin(45)\right)^{2}}+\left(\tan(45)\right)^{2}$$

Tīkina te uara \sin(45) mai i te ripanga uara pākoki.

$$11=\frac{\frac{1}{2}}{1+\left(\frac{\sqrt{2}}{2}\right)^{2}}+\left(\tan(45)\right)^{2}$$

Kia whakarewa i te \frac{\sqrt{2}}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$11=\frac{\frac{1}{2}}{1+\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}}+\left(\tan(45)\right)^{2}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 1 ki te \frac{2^{2}}{2^{2}}.

$$11=\frac{\frac{1}{2}}{\frac{2^{2}}{2^{2}}+\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}}+\left(\tan(45)\right)^{2}$$

Tā te mea he rite te tauraro o \frac{2^{2}}{2^{2}} me \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$11=\frac{\frac{1}{2}}{\frac{2^{2}+\left(\sqrt{2}\right)^{2}}{2^{2}}}+\left(\tan(45)\right)^{2}$$

Whakawehe \frac{1}{2} ki te \frac{2^{2}+\left(\sqrt{2}\right)^{2}}{2^{2}} mā te whakarea \frac{1}{2} ki te tau huripoki o \frac{2^{2}+\left(\sqrt{2}\right)^{2}}{2^{2}}.

$$11=\frac{2^{2}}{2\left(2^{2}+\left(\sqrt{2}\right)^{2}\right)}+\left(\tan(45)\right)^{2}$$

Me whakakore tahi te 2 i te taurunga me te tauraro.

$$11=\frac{2}{\left(\sqrt{2}\right)^{2}+2^{2}}+\left(\tan(45)\right)^{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$11=\frac{2}{2+2^{2}}+\left(\tan(45)\right)^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$11=\frac{2}{2+4}+\left(\tan(45)\right)^{2}$$

Tāpirihia te 2 ki te 4, ka 6.

$$11=\frac{2}{6}+\left(\tan(45)\right)^{2}$$

Whakahekea te hautanga \frac{2}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$11=\frac{1}{3}+\left(\tan(45)\right)^{2}$$

Tīkina te uara \tan(45) mai i te ripanga uara pākoki.

$$11=\frac{1}{3}+1^{2}$$

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$11=\frac{1}{3}+1$$

Tāpirihia te \frac{1}{3} ki te 1, ka \frac{4}{3}.

$$11=\frac{4}{3}$$

Me tahuri te 11 ki te hautau \frac{33}{3}.

$$\frac{33}{3}=\frac{4}{3}$$

Whakatauritea te \frac{33}{3} me te \frac{4}{3}.

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