Trigonometric problem 27th October

Question

!!(Sinx + Cosx)/Cos\^3x =atan\^3x + btan\^2x + ctanx + d!!

!!a+b+c+d = ?!!

!!a)-2!!

!!b)" "0!!

!!c)" "2!!

!!d)" "4!!

Solution

we can solve this problem using more than 2 methods.

Method 1

Given equation, !!(Sinx + Cosx)/Cos\^3x =atan\^3x + btan\^2x + ctanx + d!!

Try to convert !!LHS!! in the form of !!RHS!!.

!!=> (Sinx + Cosx)/((Cosx)xx(Cos\^2x)) =atan\^3x + btan\^2x + ctanx + d!!

!!=> ((Sinx + Cosx)/(Cosx))xx(Sec\^2x) =atan\^3x + btan\^2x + ctanx + d!!

!!=> (tanx + 1)xx(1 + tan\^2x) =atan\^3x + btan\^2x + ctanx + d!!

!!=> (tanx + tan\^3 + 1 + tan\^2) =atan\^3x + btan\^2x + ctanx + d!!

!!=> (tan\^3 + tan\^2 + tanx + 1) =atan\^3x + btan\^2x + ctanx + d!!

Equating both sides, we can get the values of !!a, b, c, d!!

!!a = 1, b = 1, c = 1, d = 1!!

Hence, !!a + b + c + d = 4!!

Method 2

!!(Sinx + Cosx)/Cos\^3x =atan\^3x + btan\^2x + ctanx + d!!

Now in the LHS add and substract !!Sin\^3x!!

We get,

!!(Sinx + Cosx + Sin\^3x - Sin\^3x )/Cos\^3x!!

!!= (Sin\^3x/Cos\^3x) + (Sinx + Cosx - Sin\^3x )/Cos\^3x!!

!!= tan\^3x + (Sinx + Cosx - (1-Cos\^2x)Sinx )/Cos\^3x!!

!!= tan\^3x + (Sinx + Cosx - Sinx + SinxCos\^2x )/Cos\^3x!!

!!= tan\^3x + (SinxCos\^2x)/Cos\^3x + (Cosx/Cos\^3x)!!

!!= tan\^3x + tanx + 1/Cos\^2x!!

!!= tan\^3x + tanx + (Sin\^2x + Cos\^2x)/Cos\^2x!!

!!= tan\^3x + tanx + (Sin\^2x/Cos\^2x) + Cos\^2x/Cos\^2x!!

!!= tan\^3x + tanx + tan\^2x + 1!!

Comparing this with !!RHS!!, we will get

!!a = 1, b = 1, c = 1, d = 1!!

Hence, !!a + b + c + d = 4!!

Method 3

By simply putting the value of !!x= 45 degree!!, we can get the value of !!a,b,c,d!!

!!(Sinx + Cosx)/Cos\^3x =atan\^3x + btan\^2x + ctanx + d!!

!!=> (1/sqrt2 + 1/sqrt2)/(1/(2sqrt2)) = a + b + c + d!!

!!=> (2/sqrt2)/(1/(2sqrt2)) = a + b + c + d!!

!!=> a + b + c + d = 4!!

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