Question

If $\sin x+cosecx=2$ then ${\sin }^{n}x+cose{c}^{n}x$ is equal to

• A. $2$
• B. $2^n$
• C. $2^{n-1}$
• D. $2^{n-2}$

Answer 1

Answer: (A)

$2$

$sinx+cosecx=2sinx+\frac{1}{sinx}=2$
Squaring both sides
${sin}^{2}x+\frac{1}{{sin}^{2}x}+2sinx.\frac{1}{sinx}=4{sin}^{2}x+\frac{1}{{sin}^{2}x}=2$
Again squaring both sides
${sin}^{4}x+\frac{1}{{sin}^{4}x}+2{sin}^{2}x.\frac{1}{{sin}^{2}x}=4{sin}^{4}x+\frac{1}{{sin}^{4}x}=2$
Similarly
${sin}^{8}x+\frac{1}{{sin}^{8}x}=2$
and
${sin}^{n}x+\frac{1}{{sin}^{n}x}=2{sin}^{n}x+{cosec}^{n}x=2$