operatorname{cosec}left[2 cot ^{-1}(5)+cos ^{ | vedclass

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

Question

(B) Let $x = 2 \cot^{-1}(5) + \cos^{-1}(\frac{4}{5})$.We know that $\cot^{-1}(5) = 	an^{-1}(\frac{1}{5})$.So,$2 \cot^{-1}(5) = 2 	an^{-1}(\frac{1}{5

Vedclass · Accepted Answer

$\frac{65}{56}$

Vedclass · Answer

(B) Let $x = 2 \cot^{-1}(5) + \cos^{-1}(\frac{4}{5})$.We know that $\cot^{-1}(5) = 	an^{-1}(\frac{1}{5})$.So,$2 \cot^{-1}(5) = 2 	an^{-1}(\frac{1}{5}) = 	an^{-1}(\frac{2(1/5)}{1-(1/5)^2}) = 	an^{-1}(\frac{2/5}{24/25}) = 	an^{-1}(\frac{2}{5} 	imes \frac{25}{24}) = 	an^{-1}(\frac{5}{12})$.Also,$\cos^{-1}(\frac{4}{5}) = 	an^{-1}(\frac{3}{4})$ because if $\cos 	heta = \frac{4}{5}$,then $	an 	heta = \frac{3}{4}$.Thus,the expression becomes $\operatorname{cosec}[	an^{-1}(\frac{5}{12}) + 	an^{-1}(\frac{3}{4})]$.Using the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1}(\frac{A+B}{1-AB})$:$	an^{-1}(\frac{5/12 + 3/4}{1 - (5/12)(3/4)}) = 	an^{-1}(\frac{(5+9)/12}{1 - 15/48}) = 	an^{-1}(\frac{14/12}{33/48}) = 	an^{-1}(\frac{14}{12} 	imes \frac{48}{33}) = 	an^{-1}(\frac{56}{33})$.Finally,$\operatorname{cosec}(	an^{-1}(\frac{56}{33}))$. Let $	heta = 	an^{-1}(\frac{56}{33})$,then $	an 	heta = \frac{56}{33}$.Since $\operatorname{cosec} 	heta = \sqrt{1 + \cot^2 	heta} = \sqrt{1 + (33/56)^2} = \sqrt{\frac{56^2 + 33^2}{56^2}} = \sqrt{\frac{3136 + 1089}{3136}} = \sqrt{\frac{4225}{3136}} = \frac{65}{56}$.

$\operatorname{cosec}\left[2 \cot ^{-1}(5)+\cos ^{-1}\left(\frac{4}{5}\right)\right]$ is equal to ..... .

Similar Questions

For $n \in N$,if $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$,then $n$ is equal to .........

If $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$,then $x =$

$\cos ^{-1}\left\{\cot \left(\sum_{i=1}^3 \cot ^{-1} i\right)\right\}=$ . . . . . . .

If $\sin^{-1} x = \theta + \beta$ and $\sin^{-1} y = \theta - \beta$,then $1 + xy = $

If $\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\frac{1}{2} \cos ^{-1} x$,then $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module