4 tan^{-1} frac{1}{5} - tan^{-1} frac{1}{239} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We use the formula $2 	an^{-1} x = 	an^{-1} \frac{2x}{1 - x^2}$.First,calculate $2 	an^{-1} \frac{1}{5}$:$2 	an^{-1} \frac{1}{5} = 	an^{-1} \

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) We use the formula $2 	an^{-1} x = 	an^{-1} \frac{2x}{1 - x^2}$.First,calculate $2 	an^{-1} \frac{1}{5}$:$2 	an^{-1} \frac{1}{5} = 	an^{-1} \frac{2(1/5)}{1 - (1/5)^2} = 	an^{-1} \frac{2/5}{24/25} = 	an^{-1} \frac{10}{24} = 	an^{-1} \frac{5}{12}$.Now,calculate $4 	an^{-1} \frac{1}{5} = 2(2 	an^{-1} \frac{1}{5}) = 2 	an^{-1} \frac{5}{12}$:$2 	an^{-1} \frac{5}{12} = 	an^{-1} \frac{2(5/12)}{1 - (5/12)^2} = 	an^{-1} \frac{10/12}{1 - 25/144} = 	an^{-1} \frac{5/6}{119/144} = 	an^{-1} \left( \frac{5}{6} 	imes \frac{144}{119} ight) = 	an^{-1} \frac{120}{119}$.Finally,evaluate $4 	an^{-1} \frac{1}{5} - 	an^{-1} \frac{1}{239}$:$	an^{-1} \frac{120}{119} - 	an^{-1} \frac{1}{239} = 	an^{-1} \left( \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} 	imes \frac{1}{239}} ight) = 	an^{-1} \left( \frac{28680 - 119}{28441 + 120} ight) = 	an^{-1} \left( \frac{28561}{28561} ight) = 	an^{-1} (1) = \frac{\pi}{4}$.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$ is equal to

Similar Questions

The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$ is:

Which kingdoms include decomposers?

Which one of the following does not correctly represent the order of property indicated against it?

If the normal to the curve $y^2 = 5x - 1$ at the point $(1, -2)$ is of the form $ax - 5y + b = 0$,then $a$ and $b$ are:

Three metal rods of the same material and identical in all respects are joined as shown in the figure. The temperatures at the ends are maintained as indicated. Assuming no loss of heat from the curved surfaces of the rods,the temperature at the junction $X$ would be ....... $^oC$

Vedclass Test Series

Exam Paper Generator

Online Exam Module