Question

if sinX+cosecx=2,then sin^19x+cosec^20x=

Answer 1

$Given, sinX+cosecX=2\\\\sinX+\frac{1}{sinX} =2\\\\\frac{sin^{2}X+1 }{sinX} =2\\\\ sin^{2}X+1=2sinX\\ sin^{2}X+1-2sinX=0\\ (sinX-1)^{2} =0\\(sinX-1)=0\\sinX=1\\sinX=sin90°\\X=90°\\\\Now,sin^{19}X+cosec^{20}X\\ (sin 90°)^{19} +(cosec90°)^{20} \\(1)^{19} +(1)^{20}\\ 1+1\\2$

Answer 2

Given :

$sin\alpha + cosec \alpha = 2$

To find :

$sin^{19}\alpha + cosec^{20}\alpha$

Solution :

Step 1

$sin\alpha + cosec\alpha = 2\\sin\alpha + \frac{1}{sin\alpha } = 2\\\\Therefore,\\\frac{1 + sin^{2}\alpha }{sin\alpha } = 2\\1 + sin^{2}\alpha = 2sin\alpha \\sin^{2}\alpha - 2sin\alpha + 1 = 0 \\(sin\alpha - 1)^{2} = 0\\Therefore , sin\alpha = 1$

We know, sine is 1 at 90°

Hence,

$\alpha$ = 90°

Step 2

Substituting this value in $sin^{19}\alpha + cosec^{20}\alpha$ , we get

= ( sin 90°)¹⁹ + ( cosec 90°)²⁰

= ( 1 )¹⁹ + ( 1 )²⁰

= 2

Final answer :

Hence, the value of $sin^{19}\alpha + cosec^{20}\alpha = 20$.