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