Question

In $A_n={\cos }^n\theta +{\sin }^n\theta ,n\in N$ and $\theta \in R$

If $A_{n-4}-A_{n-2}={\sin }^2\theta {\cos }^2\theta A_{\lambda }$ , then $\lambda$ equals

• A. $n-4$
• B. $n-8$
• C. $n-6$
• D. $n-10$

Answer 1

The correct option is C $n-6$

$A_n={\cos }^n\theta -{\sin }^n\theta$

$\Rightarrow A_{n-2}={\cos }^{n-2}\theta -{\sin }^{n-2}\theta$

$\therefore A_n-A_{n-2}={\cos }^n\theta -{\cos }^{n-2}\theta -{\sin }^n\theta +{\sin }^{n-2}\theta$

$={\cos }^{n-2}({\cos }^2\theta -1)-{\sin }^{n-2}\theta ({\sin }^2\theta -1)$

$={\cos }^2\theta {\sin }^2\theta ({\cos }^{n-4}\theta -{\sin }^{n-4}\theta )$

$={\cos }^2\theta {\sin }^2\theta A_{n-4}$

$\therefore A_{n-2}-A_{n-4}$

$=-{\sin }^2\theta {\cos }^2\theta A_{n-6}$

$\Rightarrow A_{n-4}-A_{n-2}=-{\sin }^2\theta {\cos }^2\theta A_{\lambda }$

$={\sin }^2\theta {\cos }^2\theta A_{n-6}$

$\Rightarrow \lambda =n-6$