cosA/1+sinA+1+sinA/cosA=2operatorname{sic}A eching

$$\frac { \cos A } { 1 + \sin A } + \frac { 1 + \sin A } { \cos A } = 2 \operatorname { sic } A$$

c uchun yechish

$$c=\frac{-i}{As\cos(A)}$$

$$s\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }A=\pi n_{1}+\frac{\pi }{2}\text{ and }A\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

2i hosil qilish uchun 2 va i ni ko'paytirish.

$$\frac{\cos(A)}{1+\sin(A)}+\frac{1+\sin(A)}{\cos(A)}=2iscA$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2iscA=\frac{\cos(A)}{1+\sin(A)}+\frac{1+\sin(A)}{\cos(A)}$$

Tenglama standart shaklda.

$$2iAsc=\frac{\cos(A)}{\sin(A)+1}+\frac{\sin(A)+1}{\cos(A)}$$

Ikki tarafini 2isA ga bo‘ling.

$$\frac{2iAsc}{2iAs}=\frac{2}{\cos(A)\times \left(2i\right)As}$$

2isA ga bo'lish 2isA ga ko'paytirishni bekor qiladi.

$$c=\frac{2}{\cos(A)\times \left(2i\right)As}$$

\frac{2}{\cos(A)} ni 2isA ga bo'lish.

$$c=\frac{-i}{As\cos(A)}$$