Trigonometry #01

Algebra Level 4

For $0<\phi \le \cfrac { \pi }{ 2 }$ , if $x=\sum _{ n=0 }^{ \infty }{ { cos }^{ 2n } } \phi$ , $y=\sum _{ n=0 }^{ \infty }{ { sin }^{ 2n } } \phi$ , $z=\sum _{ n=0 }^{ \infty }{ { { cos }^{ 2n }\phi .sin }^{ 2n } } \phi$ , then

Source : IIT JEE, 1992

(D)

xyz\quad =\quad x\quad +\quad yz

(C)

xyz\quad =\quad x\quad +\quad y\quad +\quad z

Both (A) and (C)

(B)

xyz\quad =\quad xy\quad +\quad z

Both (B) and (C) (A)

xyz\quad =\quad xz\quad +\quad y

Both (B) and (D) Both (A) and (B)

1 solution

Saurav Pal
Mar 1, 2015

All are infinite geometric progression with common ratio $<$ 1
$x=\cfrac { 1 }{ 1-{ cos }^{ 2 }\phi } =\cfrac { 1 }{ { sin }^{ 2 }\phi } \quad ,\quad y=\cfrac { 1 }{ { cos }^{ 2 }\phi } \quad ,\quad z=\cfrac { 1 }{ { 1-cos }^{ 2 }\phi { sin }^{ 2 }\phi } .\quad \\ Now,\quad xy+z=\frac { 1 }{ { sin }^{ 2 }\phi { cos }^{ 2 }\phi (1-{ sin }^{ 2 }\phi { cos }^{ 2 }\phi ) } =xyz.\quad \quad \\ x+y=\cfrac { { sin }^{ 2 }\phi +{ cos }^{ 2 }\phi }{ { sin }^{ 2 }\phi { cos }^{ 2 }\phi } =xy.\quad \\ \therefore \quad x+y+z=xyz.\quad$

Trigonometry #01

1 solution

0 pending reports