Question

If $\sin x+\cos ecx=2$ then ${\sin }^{n}x+\cos e{c}^{n}x=?$

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

Answer 1

The correct option is D $2$

$sinx+cosecx=2$

squaring both side

$\Rightarrow (sinx+cosecx{)}^2=2^2$

$\Rightarrow si{n}^{2}x+cose{c}^{2}x+2sinxcosecx=2^2$

$\Rightarrow x+cose{c}^{2}x+2=2^2$ (as sin x cosec x=1)

$\Rightarrow si{n}^{2}x+cose{c}^{2}x=2^2-2$

$\Rightarrow si{n}^{2}x+cose{c}^{2}x=2$

again squaring both side we get

$\Rightarrow si{n}^{4}x+cose{c}^{4}x=2$

Similarly we can say that

$\Rightarrow si{n}^{n}x+cose{c}^{n}x=2$ $\text{(such that n is even)}$