Question

solve this question ​

Answer 1

TO FIND :–

$\\ \: \: \: \left|\begin{array}{cc}\sf cos\:x & \sf\:sin\:x\\\sf\:-\:sin\:x & \sf\:cos\:x\end{array}\right| = ?$

SOLUTION :–

• Let Determinant be –

$\\ \: \implies \: A = \left|\begin{array}{cc}\sf cos\:x & \sf\:sin\:x\\\sf\:-\:sin\:x & \sf\:cos\:x\end{array}\right|$

• We know that –

$\\ \: \dashrightarrow \: A = \left|\begin{array}{cc}\sf a_{11} & \sf a_{12}\\\sf\:a_{21} & \sf a_{22}\end{array}\right|$

• Then –

$\\ \: \dashrightarrow \sf A =a_{11}.a_{22} -a_{21}.a_{12}$

• Here –

$\\ \sf \: \: { \huge{.}} \: \: a_{11} = \cos x$

$\\ \sf \: \: { \huge{.}} \: \: a_{12} = \sin x$

$\\ \sf \: \: { \huge{.}} \: \: a_{21} = - \sin x$

$\\ \sf \: \: { \huge{.}} \: \: a_{22} = \cos x$

• So that –

$\\ \: \implies \sf A = ( \cos x).(\cos x) - ( \sin x)( - \sin x)$

$\\ \: \implies \sf A = \cos^{2} x + \sin^{2} x$

$\\ \: \: \: \sf \because \: \: \: \cos^{2} x + \sin^{2} x = 1$

$\\ \: \: \: \sf \therefore \: \: \: \large{ \boxed{ \sf A= 1}}$

$\\ \rule{220}{2}$