Question

If $\tan {\theta }_1,\tan {\theta }_2,\tan {\theta }_3,\tan {\theta }_4$ are the roots of the equation $x^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =0$ then $\tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)=$

• A. $\sin \beta$
• B. $\cos \beta$
• C. $\tan \beta$
• D. $\cot \beta$

Answer 1

Answer: (D)

$\cot \beta$

From the given equation we get
$S_1=\tan {\theta }_1+\tan {\theta }_2+\tan {\theta }_3+\tan {\theta }_4=\sin 2\beta$
$S_2=\sum \tan {\theta }_1\tan {\theta }_2=\cos 2\beta$
$S_3=\sum \tan {\theta }_1\tan {\theta }_2\tan {\theta }_3=\cos \beta$
and $S_4=\tan {\theta }_1\tan {\theta }_2\tan {\theta }_3\tan {\theta }_4=-\sin \beta$
Now $\tan ({\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4)=\frac{S_1-S_3}{1-S_2+S_4}$.
$=\frac{\sin 2\beta -\cos \beta }{1-{\cos }^2\beta -\sin \beta }=\frac{\cos \beta (2\sin \beta -1)}{\sin \beta (2\sin \beta -1)}=cot\beta$.