wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate the determinant
(i) cosθsinθsinθcosθ
(ii) x2x+1x1x+1x+1


Open in App
Solution

(i) cosθsinθsinθcosθ
=cosθ(cosθ)(sinθ)(sinθ)
=cos2θ+sin2θ
=1
[cos2θ+sin2θ=1]

(ii) x2x+1x1x+1x+1
=(x2x+1)(x+1)(x+1)(x1)
=(x2x+1)(x+1)(x21)
=[(x2x+1)x+(x2x+1)1](x21)
=(x3x2+x+x2x+1)(x21)
=x3+1x2+1
=x3x2+2

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon