frac{1-sin^{2}45^{circ}}{1+sin^245^circ}+tan^245^circ eching

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

Baholash

$$\frac{4}{3}\approx 1,333333333$$

Yechim qadamlari

Trigonometrik qiymatlar jadvaldan \sin(45) qiymatini oling.

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

\frac{\sqrt{2}}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

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

\sqrt{2} kvadrati – 2.

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

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

\frac{2}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

\frac{1}{2} olish uchun 1 dan \frac{1}{2} ni ayirish.

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

Trigonometrik qiymatlar jadvaldan \sin(45) qiymatini oling.

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

\frac{\sqrt{2}}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

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

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 1 ni \frac{2^{2}}{2^{2}} marotabaga ko'paytirish.

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

\frac{2^{2}}{2^{2}} va \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

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

\frac{1}{2} ni \frac{2^{2}+\left(\sqrt{2}\right)^{2}}{2^{2}} ga bo'lish \frac{1}{2} ga k'paytirish \frac{2^{2}+\left(\sqrt{2}\right)^{2}}{2^{2}} ga qaytarish.

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

Surat va maxrajdagi ikkala 2 ni qisqartiring.

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

\sqrt{2} kvadrati – 2.

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

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

6 olish uchun 2 va 4'ni qo'shing.

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

\frac{2}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Trigonometrik qiymatlar jadvaldan \tan(45) qiymatini oling.

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

2 daraja ko‘rsatkichini 1 ga hisoblang va 1 ni qiymatni oling.

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

\frac{4}{3} olish uchun \frac{1}{3} va 1'ni qo'shing.

$$\frac{4}{3}$$