Question

If $\tan \left(\theta \right)-cot\left(\theta \right)=a$ and $\sin \left(\theta \right)+\cos \left(\theta \right)=b$, then ${\left(b^2-1\right)}^2\left(a^2+4\right)$ is equal to

• A. $2$
• B. $\pm 4$
• C. $-4$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for the correct option

Given: $\tan \left(\theta \right)-cot\left(\theta \right)=a$ and $\sin \left(\theta \right)+\cos \left(\theta \right)=b$

$$\begin{gathered} \therefore {\left(b^2-1\right)}^2\left(a^2+4\right)={\left[{\left(\sin \left(\theta \right)+\cos \left(\theta \right)\right)}^2-1\right]}^2\left[\left(\tan \left(\theta \right)-\cot {\left(\theta \right)}^2+4\right)\right] \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)={\left[\left({\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)+2\sin \left(\theta \right)\cos \left(\theta \right)\right)-1\right]}^2\left[{\left(\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}-\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}\right)}^2+4\right] \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)={\left[1+2\sin \left(\theta \right)\cos \left(\theta \right)-1\right]}^2\left[{\left(\frac{{\sin }^2\left(\theta \right)-{\cos }^2\left(\theta \right)}{\sin \left(\theta \right)\cos \left(\theta \right)}\right)}^2+4\right] \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)=\left[4{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)\right]\left[\left(\frac{{\sin }^4\left(\theta \right)+{\cos }^4\left(\theta \right)-2{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)+4{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)}{{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)}\right)\right] \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)=\left[4\right]\left({\sin }^4\left(\theta \right)+{\cos }^4\left(\theta \right)+2{\sin }^2\left(\theta \right){\cos }^2\left(\theta \right)\right) \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)=\left[4\right]{\left({\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)\right)}^2 \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)=\left[4\right]{\left(1\right)}^2\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \\ \Rightarrow {\left(b^2-1\right)}^2\left(a^2+4\right)=4 \end{gathered}$$

Hence, option D is correct.