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Question: If \(\tan x=\dfrac{2b}{a-c}\),\((a\ne c)\), \(y=a{{\cos }^{2}}x+2b\sin x\cos x+c{{\sin }^{2}}x\) and...

If tan⁡x=2ba−c\tan x=\dfrac{2b}{a-c},(a≠c)(a\ne c), y=acos⁡2x+2bsin⁡xcos⁡x+csin⁡2xy=a{{\cos }^{2}}x+2b\sin x\cos x+c{{\sin }^{2}}x and z=asin⁡2x−2bsin⁡xcos⁡x+ccos⁡2xz=a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x, then
a). y=zy=z
b). y+z=a+cy+z=a+c
c). y−z=a−cy-z=a-c
d). y−z=(a−c)2+4b2y-z={{\left( a-c \right)}^{2}}+4{{b}^{2}}

Explanation

Solution

We are asked to find the relation between y, z, a, b and c; with the given value of ‘y’ and ‘z’ in terms of trigonometric function and a, b, c. so we will simply find the value of ′y−z′'y-z' as it is given in two options, if we got the value same as given in option then it is right, otherwise we will move to other options.

Complete step-by-step solution:
Moving ahead with the question in step wise manner we had,
y=acos⁡2x+2bsin⁡xcos⁡x+csin⁡2xy=a{{\cos }^{2}}x+2b\sin x\cos x+c{{\sin }^{2}}x
z=asin⁡2x−2bsin⁡xcos⁡x+ccos⁡2xz=a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}xand
tan⁡x=2ba−c\tan x=\dfrac{2b}{a-c},
So let us find out the value of ′y−z′'y-z' as given in the two options, so we will get;
y−z=acos⁡2x+2bsin⁡x+csin⁡2x−(asin⁡2x−2bsin⁡xcos⁡x+ccos⁡2x)y-z=a{{\cos }^{2}}x+2b\sin x+c{{\sin }^{2}}x-\left( a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x \right)
On further simplifying using the basic Mathematical operations and trigonometric identities, we will get;
⇒y−z=acos⁡2x+2bsin⁡xcos⁡x+csin⁡2x−(asin⁡2x−2bsin⁡xcos⁡x+ccos⁡2x) ⇒y−z=a(cos⁡2x−sin⁡2x)+2b(2sin⁡xcos⁡x)−c(cos⁡2x−sin⁡2x) \begin{aligned} &\Rightarrow y-z=a{{\cos }^{2}}x+2b\sin x\cos x+c{{\sin }^{2}}x-\left( a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x \right) \\\ &\Rightarrow y-z=a\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)+2b\left( 2\sin x\cos x \right)-c\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \\\ \end{aligned}
As we know that (cos⁡2x−sin⁡2x)\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) is equal to cos⁡2x\cos 2xand2sin⁡xcos⁡x2\sin x\cos x equal to sin⁡2x\sin 2x, so by using the same identity in above equation we will get;
⇒y−z=acos⁡2x+2bsin⁡2x−ccos⁡2x ⇒y−z=cos⁡2x(a−c)+2bsin⁡2x \begin{aligned} &\Rightarrow y-z=a\cos 2x+2b\sin 2x-c\cos 2x \\\ &\Rightarrow y-z=\cos 2x\left( a-c \right)+2b\sin 2x \\\ \end{aligned}
Since in the question we are given with the value of tan⁡x\tan x in terms of a, b, c. so let us reduce the above trigonometric equation of tan⁡x\tan x. So as we know that cos⁡2x=1−tan⁡2x1+tan⁡2x\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} and sin⁡2x=2tan⁡x1+tan⁡2x\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}, by using the same identity in above equation we will get;
y−z=(1−tan⁡2x1+tan⁡2x)(a−c)+2b(2tan⁡x1+tan⁡2x)y-z=\left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)\left( a-c \right)+2b\left( \dfrac{2\tan x}{1+{{\tan }^{2}}x} \right)
On further simplifying it by taking the LCM we will get;
y−z=(1−tan⁡2x)(a−c)+2b(2tan⁡x)1+tan⁡2xy-z=\dfrac{\left( 1-{{\tan }^{2}}x \right)\left( a-c \right)+2b\left( 2\tan x \right)}{1+{{\tan }^{2}}x}
As according to the given information we know that tan⁡x=2ba−c\tan x=\dfrac{2b}{a-c}, so replace tan⁡x\tan x with the given information, so we will get;
y−z=(1−(2ba−c)2)(a−c)+2b(2(2ba−c))1+(2ba−c)2y-z=\dfrac{\left( 1-{{\left( \dfrac{2b}{a-c} \right)}^{2}} \right)\left( a-c \right)+2b\left( 2\left( \dfrac{2b}{a-c} \right) \right)}{1+{{\left( \dfrac{2b}{a-c} \right)}^{2}}}
On solving we will get;

&\Rightarrow y-z=\dfrac{\dfrac{\left( {{\left( a-c \right)}^{2}}-4{{b}^{2}} \right)+8{{b}^{2}}}{\left( a-c \right)}}{\dfrac{{{\left( a-c \right)}^{2}}+4{{b}^{2}}}{{{\left( a-c \right)}^{2}}}} \\\ &\Rightarrow y-z=\dfrac{\left( {{a}^{2}}+{{c}^{2}}-2ac+4{{b}^{2}} \right)\left( a-c \right)}{{{\left( a-c \right)}^{2}}+4{{b}^{2}}} \\\ \end{aligned}$$ So we got; $$\begin{aligned} &\Rightarrow y-z=\dfrac{{{\left( a-c \right)}^{2}}+4{{b}^{2}}\left( a-c \right)}{{{\left( a-c \right)}^{2}}+4{{b}^{2}}} \\\ &\Rightarrow y-z=a-c \\\ \end{aligned}$$ So on simplifying we got $$y-z=a-c$$ which is equal to option ‘c’. **Hence the correct answer is option ‘c’ i.e. $$y-z=a-c$$.** **Note:** Motive behind first finding the value of $'y-z'$ is only because of its presence in more than one option, other than this we can go with $'y+z'$, then we will get the relation which will be not equal to given option, so then also correct answer will be option ‘c’.