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 $z=a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x$, then
a). $y=z$
b). $y+z=a+c$
c). $y-z=a-c$
d). \[y-z={{\left( a-c \right)}^{2}}+4{{b}^{2}}\]

Answer 1

Hint: 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 $'y-z'$ 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,
$y=a{{\cos }^{2}}x+2b\sin x\cos x+c{{\sin }^{2}}x$
$z=a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x$and
$\tan x=\dfrac{2b}{a-c}$,
So let us find out the value of $'y-z'$ as given in the two options, so we will get;
$y-z=a{{\cos }^{2}}x+2b\sin x+c{{\sin }^{2}}x-\left( a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x \right)$
On further simplifying using the basic Mathematical operations and trigonometric identities, we will get;
$\begin{align}
  &\Rightarrow y-z=a{{\cos }^{2}}x+2b\sin x\cos x+c{{\sin }^{2}}x-\left( a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x \right) \\
 &\Rightarrow y-z=a\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)+2b\left( 2\sin x\cos x \right)-c\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \\
\end{align}$
As we know that $\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$ is equal to $\cos 2x$and$2\sin x\cos x$ equal to $\sin 2x$, so by using the same identity in above equation we will get;
$\begin{align}
  &\Rightarrow y-z=a\cos 2x+2b\sin 2x-c\cos 2x \\
 &\Rightarrow y-z=\cos 2x\left( a-c \right)+2b\sin 2x \\
\end{align}$
Since in the question we are given with the value of $\tan x$ in terms of a, b, c. so let us reduce the above trigonometric equation of $\tan x$. So as we know that $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ and $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$, by using the same identity in above equation we will get;
$y-z=\left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)\left( a-c \right)+2b\left( \dfrac{2\tan x}{1+{{\tan }^{2}}x} \right)$
On further simplifying it by taking the LCM we will get;
$y-z=\dfrac{\left( 1-{{\tan }^{2}}x \right)\left( a-c \right)+2b\left( 2\tan x \right)}{1+{{\tan }^{2}}x}$
As according to the given information we know that $\tan x=\dfrac{2b}{a-c}$, so replace $\tan x$ with the given information, so we will get;
\[y-z=\dfrac{\left( 1-{{\left( \dfrac{2b}{a-c} \right)}^{2}} \right)\left( a-c \right)+2b\left( 2\left( \dfrac{2b}{a-c} \right) \right)}{1+{{\left( \dfrac{2b}{a-c} \right)}^{2}}}\]
On solving we will get;
\[\begin{align}
  &\Rightarrow y-z=\dfrac{\dfrac{\left( {{\left( a-c \right)}^{2}}-4{{b}^{2}} \right)+8{{b}^{2}}}{\left( a-c \right)}}{\dfrac{{{\left( a-c \right)}^{2}}+4{{b}^{2}}}{{{\left( a-c \right)}^{2}}}} \\
 &\Rightarrow y-z=\dfrac{\left( {{a}^{2}}+{{c}^{2}}-2ac+4{{b}^{2}} \right)\left( a-c \right)}{{{\left( a-c \right)}^{2}}+4{{b}^{2}}} \\
\end{align}\]
So we got;
\[\begin{align}
  &\Rightarrow y-z=\dfrac{{{\left( a-c \right)}^{2}}+4{{b}^{2}}\left( a-c \right)}{{{\left( a-c \right)}^{2}}+4{{b}^{2}}} \\
 &\Rightarrow y-z=a-c \\
\end{align}\]
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’.