Question

$If sinx + cosecx = 2.Find sin^{19}x + cosec^{20}x$
​

Answer 1

sin x + cosec x = 2

sin x + 1 /sin x = 2

sin^2 x + 1 /sin x = 2

sin^2 x + 1 = 2sin x

sin^2 x + 1 - 2 sinx = 0

(sin x - 1)^2 = 0

(sin x - 1 ) = 0

sin x = 1

sin x = sin 90°

x = 90°

sin^19x + cosec^20x

(sin 90°)^19 + (cosec 90°)^20

(1)^19 + (1)^19

1 + 1

= 2

Therefore, sin^19 x + cosec^20 x = 2

Answer 2

Hope it helps and solves your problem