If cotθ=4, find the value of 5 sin θ−3 cos θ 5 sin θ+3 cos θ
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let the angle theta=A
because I have no symbol of theta
cotA= Base/perpendicular
cotA= 4/1
by Pythagoreans theorem,
(H)^2=(B)^2 + (P)^2
H=(4)^2+(1)^2 under square root
H=2
sinA= perpendicular/hypotenuse
sin A= 1/2
cos A=Base/hypotenuse
cos A=4/2=2
ATQ
(5 Sin A - 3 cos A)(5sin A + 3 cos A)
=( 5×1/2 - 3 ×2) ( 5×1/2 + 3×2)
=(5-12/2)(5/2+6)
=(5-12/2)(5+12/2)
=( -7/2) (17/2)
=( -119/4)
Hope u got it!
because I have no symbol of theta
cotA= Base/perpendicular
cotA= 4/1
by Pythagoreans theorem,
(H)^2=(B)^2 + (P)^2
H=(4)^2+(1)^2 under square root
H=2
sinA= perpendicular/hypotenuse
sin A= 1/2
cos A=Base/hypotenuse
cos A=4/2=2
ATQ
(5 Sin A - 3 cos A)(5sin A + 3 cos A)
=( 5×1/2 - 3 ×2) ( 5×1/2 + 3×2)
=(5-12/2)(5/2+6)
=(5-12/2)(5+12/2)
=( -7/2) (17/2)
=( -119/4)
Hope u got it!
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