Question

sin x + cosec x =2 , sin ^8 + cosec ^ 8 x=​

Answer 1

Answer:

Step-by-step explanation:

sinx+1/sinx=2

(sinx+1/sinx)^2 =sin^ 2 x + 2 + 1/sin^2 x =4

sin^ 2 x + 1/sin^2 x =2                !!! \:

sin^ 8 x + 1/ sin^8 x =2

Answer 2

Answer:

$Sin^{8}x + Cosec^{8} x$ = 2

Step-by-step explanation:

Given : sin x + cosec x =2

To find : $Sin^{8}x + Cosec^{8} x$

Solution:

Sin x + cosec x =2

Sin x + $\frac{1}{Sinx}$ = 2 (Sin x is the reciprocal of cosec x)

$\frac{Sin^{2}x + 1 }{Sinx}$ = 2

$Sin^{2} x$ + 1 = 2 Sin x

$Sin^{2} x$ + 1 - 2 sin x = 0

Using algebraic identity $(x - y)^{2} = x^{2} - 2xy + y^{2}$-

$(Sinx-1)^{2}$ = 0

Sin x - 1 = 0

Sin x = 1

Now,

$Sin^{8}x + Cosec^{8} x$

$= 1^{8} + (\frac{1}{Sin x} )^{8} \\= 1^{8} + 1^{8}$

= 1 + 1

= 2.

$Sin^{8}x + Cosec^{8} x$ = 2

#SPJ3