Question

If sinx + cosecx = -2, then for a positive integer n, sin^n.x + cosec^n.x is
a) 2 b) -2
c) -2 if n is odd and 2{if n is even} d) none

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\sin^n\:x\:+\:cosec^n\:x\:=\:-\:2\:}}}$

Option b) - 2

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\sin\:x\:+\:cosec\:x\:=\:-\:2}$

We have to find the value of

$\displaystyle{\sf\:\sin^n\:x\:+\:cosec^n\:x}$ for a positive integer n.

Now,

$\displaystyle{\sf\:\sin\:x\:+\:cosec\:x\:=\:-\:2}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:cosec\:x\:=\:\dfrac{1}{\sin\:x}\:}}}$

$\displaystyle{\implies\sf\:\sin\:x\:+\:\dfrac{1}{\sin\:x}\:=\:-\:2}$

$\displaystyle{\implies\sf\:\dfrac{\sin^2\:x\:+\:1}{\sin\:x}\:=\:-\:2}$

$\displaystyle{\implies\sf\:\sin^2\:x\:+\:1\:=\:-\:2\:\sin\:x}$

$\displaystyle{\implies\sf\:\sin^2\:x\:+\:2\:\sin\:x\:+\:1\:=\:0}$

$\displaystyle{\implies\sf\:\sin^2\:x\:+\:\sin\:x\:+\:\sin\:x\:+\:1\:=\:0}$

$\displaystyle{\implies\sf\:\sin\:x\:(\:\sin\:x\:+\:1\:)\:+\:1\:(\:\sin\:x\:+\:1\:)\:=\:0}$

$\displaystyle{\implies\sf\:(\:\sin\:x\:+\:1\:)\:(\:\sin\:x\:+\:1\:)\:=\:0}$

$\displaystyle{\implies\sf\:(\:\sin\:x\:+\:1\:)\:=\:0}$

$\displaystyle{\implies\sf\:\sin\:x\:+\:1\:=\:0}$

$\displaystyle{\implies\sf\:\sin\:x\:=\:-\:1}$

$\displaystyle{\implies\sf\:-\:\sin\:x\:=\:1}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\sin\:(\:-\:x\:)\:=\:-\:\sin\:(\:x\:)\:}}}$

$\displaystyle{\sf\:\sin\:\left(\:\dfrac{\pi}{2}\:\right)\:=\:1}$

$\displaystyle{\implies\sf\:\sin\:(\:-\:x\:)\:=\:\sin\:\left(\:\dfrac{\pi}{2}\:\right)}$

$\displaystyle{\implies\sf\:\arcsin\:(\:\sin\:(\:-\:x\:)\:)\:=\:\arcsin\:\left(\:\sin\:\left(\:\dfrac{\pi}{2}\:\right)\:\right)}$

$\displaystyle{\implies\sf\:-\:x\:=\:\dfrac{\pi}{2}}$

$\displaystyle{\implies\:\boxed{\orange{\sf\:x\:=\:-\:\dfrac{\pi}{2}\:}}}$

Now,

$\displaystyle{\sf\:\sin\:\left(\:-\:\dfrac{\pi}{2}\:\right)\:=\:-\:1}$

$\displaystyle{\therefore\:\boxed{\green{\sf\:\sin\:x\:=\:-\:1}}}$

$\displaystyle{\sf\:cosec\:\left(\:-\:\dfrac{\pi}{2}\:\right)\:=\:\dfrac{1}{\sin\:\left(\:-\:\dfrac{\pi}{2}\:\right)}}$

$\displaystyle{\implies\sf\:cosec\:\left(\:-\:\dfrac{\pi}{2}\:\right)\:=\:\dfrac{1}{-\:1}}$

$\displaystyle{\implies\sf\:cosec\:\left(\:-\:\dfrac{\pi}{2}\:\right)\:=\:-\:1}$

$\displaystyle{\therefore\:\boxed{\purple{\sf\:cosec\:x\:=\:-\:1}}}$

Now, we have to find the value of

$\displaystyle{\sf\:\sin^n\:x\:+\:cosec^n\:x}$

$\displaystyle{\implies\sf\:(\:\sin\:x\:)^n\:+\:(\:cosec\:x\:)^n}$

$\displaystyle{\implies\sf\:(\:-\:1\:)^n\:+\:(\:-\:1\:)^n}$

For a positive integer n, ( - 1 )ⁿ is always - 1.

$\displaystyle{\implies\sf\:(\:-\:1\:)\:+\:(\:-\:1\:)}$

$\displaystyle{\implies\sf\:-\:1\:-\:1}$

$\displaystyle{\implies\sf\:-\:2}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\sin^n\:x\:+\:cosec^n\:x\:=\:-\:2\:}}}}$

Answer 2

Step-by-step explanation:

$\blue{ \text{Since, \( \tt \sin x+\operatorname{cosec} x=2 \) }}$

$\color{red} \pmb{\[ \begin{aligned} \Rightarrow & \sin x+\frac{1}{\sin x} =2 \\ \Rightarrow & \sin ^{2} x-2 \sin x+1 =0 \\ \Rightarrow &(\sin x-1)^{2} =0 \\ \Rightarrow & \sin x =1 \\ \text { Now, } \quad & \sin ^{n} x+\operatorname{cosec}^{n} x &=\sin ^{n} x+\frac{1}{\sin ^{n} x} \\ &=1+1=2 \end{aligned} \]}$