4 tan ^{-1} frac{1}{5}-tan ^{-1} frac{1}{70}+ | vedclass

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

Question

(C) We use the formula $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.First,calculate $2 	an ^{-1} \frac{1}{5} = 	an ^{-1} \left( \fr

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) We use the formula $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.First,calculate $2 	an ^{-1} \frac{1}{5} = 	an ^{-1} \left( \frac{2/5}{1-1/25} ight) = 	an ^{-1} \left( \frac{2/5}{24/25} ight) = 	an ^{-1} \frac{5}{12}$.Then,$4 	an ^{-1} \frac{1}{5} = 2 	an ^{-1} \frac{5}{12} = 	an ^{-1} \left( \frac{2(5/12)}{1-(5/12)^2} ight) = 	an ^{-1} \left( \frac{5/6}{1-25/144} ight) = 	an ^{-1} \left( \frac{5/6}{119/144} ight) = 	an ^{-1} \frac{120}{119}$.Now,evaluate $-	an ^{-1} \frac{1}{70} + 	an ^{-1} \frac{1}{99} = 	an ^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} ight) = 	an ^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} ight) = 	an ^{-1} \left( \frac{-29}{6931} ight) = -	an ^{-1} \frac{1}{239}$.Finally,$	an ^{-1} \frac{120}{119} - 	an ^{-1} \frac{1}{239} = 	an ^{-1} \left( \frac{120/119 - 1/239}{1 + (120/119)(1/239)} ight) = 	an ^{-1} \left( \frac{(28680-119)/(119 	imes 239)}{(28441+120)/(119 	imes 239)} ight) = 	an ^{-1} \left( \frac{28561}{28561} ight) = 	an ^{-1} 1 = \frac{\pi}{4}$.

$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{70}+\tan ^{-1} \frac{1}{99}=$

Similar Questions

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1,$ then find the value of $x$.

If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}$,then $\sin^{-1} A + \tan^{-1} B + \sec^{-1} C$ is equal to

$\cot ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) = $ . . . . . .

Let $S = \{x \in R : 0 < x < 1 \text{ and } 2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$. If $n(S)$ denotes the number of elements in $S$,then:

$\sin^{-1} x + \cos^{-1} x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module