If f(α ) [cos α&sin α -sin α &cos α ] and if α, β, γ, are angle of a triangle, then f (α).f(β).f(γ) equals

Question

If f(α ) [cos α&sin α -sin α &cos α ] and if α, β, &gamma;, are angle of a triangle, then f (α).f(β).f(&gamma;) equals (A) I2 (B) -I2 (C) 0 (D) None of these

Tardigrade Admin · Accepted Answer

Hence f(α)f(β)=[cos α&sin α -sin α &cos α ][cos β&sin β -sin α β&cos β] [cos α cos β-sin α sin β&cos α sin β+sin α cos β -sin α cos β-cos α sin β&-sin α sin β+cos α cos β] =[cos(α+β)&sin(α+β) -sin(α+β)&cos(α+β)] similarly f(α)f(β)f(&gamma;)=[cos(α+β+&gamma;)&sin(α+β+&gamma;) -sin(α+β+&gamma; )&cos(α+β+&gamma; )] =[cos π &sin π -sin π &cos π ] as α+β +&gamma; =π =[-1&0 0&-1]=-[1&0 0&1]=-I2

Q. If $f (α) [cos α - s in α s in α cos α]$ and if $α, β, γ,$ are angle of a triangle, then $f (α) . f (β) . f (γ)$ equals

$I_{2}$

$- I_{2}$

$0$

None of these

Q. If f(α)[cosα−sinα​sinαcosα​] and if α,β,γ, are angle of a triangle, then f(α).f(β).f(γ) equals

I2​

−I2​

0

None of these

Q. If $f (α) [cos α - s in α s in α cos α]$ and if $α, β, γ,$ are angle of a triangle, then $f (α) . f (β) . f (γ)$ equals

$I_{2}$

$- I_{2}$

$0$