Question

(1+sinx+icosx/1+sinx-icosx)^n

Answer 1

Step-by-step explanation:

(1+sinx)+(icosx)

(1+sinx)-(icosx)

(1+sinx)+(icosx)[(1+sinx)+(icosx)]

(1+sinx)-(icosx)[(1+sinx)+(icosx)]

[(1+sinx)+(icosx)]²

[(1+sinx)²-(icosx)²]

(1+sinx)²+i²cos²x+2(1+sinx)(icosx)

[1+sin²x+2sinx-i²cos²x]

1+sin²x+2sinx-cos²x+2(icosx+icosxsinx)

[1+sin²x+2sinx+cos²x]