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Question Number 81873 by zainal tanjung last updated on 16/Feb/20

 determinant (((sin^2 x),(cos^2 x),1),((cos^2 x),(sin^2 x),1),((−10),(  12),2))=

$$\begin{vmatrix}{\mathrm{sin}^{\mathrm{2}} {x}}&{\mathrm{cos}^{\mathrm{2}} {x}}&{\mathrm{1}}\\{\mathrm{cos}^{\mathrm{2}} {x}}&{\mathrm{sin}^{\mathrm{2}} {x}}&{\mathrm{1}}\\{−\mathrm{10}}&{\:\:\mathrm{12}}&{\mathrm{2}}\end{vmatrix}= \\ $$

Commented by john santu last updated on 16/Feb/20

sin^2 x(2sin^2 x−12)−cos^2 x(2cos^2 x+10)−(12cos^2 x+10sin^2 x)

$$\mathrm{sin}\:^{\mathrm{2}} {x}\left(\mathrm{2sin}\:^{\mathrm{2}} {x}−\mathrm{12}\right)−\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{10}\right)−\left(\mathrm{12cos}\:^{\mathrm{2}} {x}+\mathrm{10sin}\:^{\mathrm{2}} {x}\right) \\ $$

Answered by MJS last updated on 16/Feb/20

=2(sin^4  x −sin^2  x −cos^4  x +cos^2  x)=       [cos x =(√(1−sin^2  x))]  =0

$$=\mathrm{2}\left(\mathrm{sin}^{\mathrm{4}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{cos}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{2}} \:{x}\right)= \\ $$$$\:\:\:\:\:\left[\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}\right] \\ $$$$=\mathrm{0} \\ $$

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