Question

If $f\left(\theta \right)=\cos {\theta }_1\cdot \cos {\theta }_2\cdot \cos {\theta }_3.....\cos {\theta }_n$. Also $\frac{d{\theta }_1}{d\theta }=\frac{d{\theta }_2}{d\theta }=.....=\frac{d{\theta }_n}{d\theta }=1$, then $\{\tan {\theta }_1+\tan {\theta }_2+\tan {\theta }_3+....+\tan {\theta }_n\}=$

• A. $-\left\{\frac{f^{\prime }\left(\theta \right)}{f\left(\theta \right)}\right\}$
• B. $\left\{\frac{f^{\prime \prime }\left(\theta \right)}{f\left(\theta \right)}\right\}$
• C. $-\left\{\frac{f\left(\theta \right)}{f^{\prime }\left(\theta \right)}\right\}$
• D. none of these

Answer 1

Answer: (A)

$-\left\{\frac{f^{\prime }\left(\theta \right)}{f\left(\theta \right)}\right\}$

$f\left(\theta \right)=cos{\theta }_1\cdot cos{\theta }_2\cdot cos{\theta }_3.....cos{\theta }_n$
Taking log on both the sides,
$logf\left(\theta \right)=log\left(cos{\theta }_1\right)+log\left(cos{\theta }_2\right)+.....+log\left(cos{\theta }_n\right)$
On differentiating both the sides w.r.t. $\theta$, we get
$\frac{1}{f\left(\theta \right)}\cdot f^{\prime }\left(\theta \right)=\frac{1}{cos{\theta }_1}\cdot (-sin{\theta }_1)+\frac{1}{cos{\theta }_2}\cdot (-sin{\theta }_2)+.....+\frac{1}{cos{\theta }_n}\cdot (-sin{\theta }_n)$
Hence, $(tan{\theta }_1+tan{\theta }_2+.....+tan{\theta }_n)=-\left\{\frac{f^{\prime }\left(\theta \right)}{f\left(\theta \right)}\right\}$