Question

If $\tan \left(x\right)=\frac{2b}{a-c},(a\ne c),y=a{\cos }^2\left(x\right)+2b\sin \left(x\right)\cos \left(x\right)+c{\sin }^2\left(x\right)$ and $z=a{\sin }^2\left(x\right)-2b\sin \left(x\right)\cos \left(x\right)+c{\cos }^2\left(x\right),$then

• A. $y=z$
• B. $y+z=a+c$
• C. $y-z=a-c$
• D. $y-z={(a-c)}^2+4b^2$

Answer 1

The correct option is B

$y+z=a+c$

Explanation for correct option

Given: $\tan \left(x\right)=\frac{2b}{a-c},(a\ne c),y=a{\cos }^2\left(x\right)+2b\sin \left(x\right)\cos \left(x\right)+c{\sin }^2\left(x\right)$ and $z=a{\sin }^2\left(x\right)-2b\sin \left(x\right)\cos \left(x\right)+c{\cos }^2\left(x\right)$

$y=a{\cos }^2\left(x\right)+2b\sin \left(x\right)\cos \left(x\right)+c{\sin }^2\left(x\right)$

$z=a{\sin }^2\left(x\right)-2b\sin \left(x\right)\cos \left(x\right)+c{\cos }^2\left(x\right)$

$$\begin{gathered} \therefore y+z=a{\cos }^2\left(x\right)+2b\sin \left(x\right)\cos \left(x\right)+c{\sin }^2\left(x\right)+a{\sin }^2\left(x\right)-2b\sin \left(x\right)\cos \left(x\right)+c{\cos }^2\left(x\right) \\ \Rightarrow y+z=a\left({\sin }^2\right(x)+{\cos }^2(x\left)\right)+\left(2b\sin \left(x\right)\cos \left(x\right)-2b\sin \left(x\right)\cos (x\right))+c({\sin }^2\left(x\right)+{\cos }^2\left(x\right)) \\ \Rightarrow y+z=a+c\left[\because {\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1\right] \end{gathered}$$

Hence, option B is correct.