cos^{-1}left(frac{15}{17}right) + 2tan^{-1}le | vedclass

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

Question

(D) આપેલ પદાવલિ: $\cos^{-1}\left(\frac{15}{17}ight) + 2	an^{-1}\left(\frac{1}{5}ight)$.પ્રથમ,$2	an^{-1}\left(\frac{1}{5}ight)$ ને $\cos^{-1}$

Vedclass · Accepted Answer

આમાંથી કોઈ નહીં

Vedclass · Answer

(D) આપેલ પદાવલિ: $\cos^{-1}\left(\frac{15}{17}ight) + 2	an^{-1}\left(\frac{1}{5}ight)$.પ્રથમ,$2	an^{-1}\left(\frac{1}{5}ight)$ ને $\cos^{-1}$ સ્વરૂપમાં ફેરવવા માટે સૂત્ર $2	an^{-1}(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$ નો ઉપયોગ કરો:$2	an^{-1}\left(\frac{1}{5}ight) = \cos^{-1}\left(\frac{1 - (1/5)^2}{1 + (1/5)^2}ight) = \cos^{-1}\left(\frac{1 - 1/25}{1 + 1/25}ight) = \cos^{-1}\left(\frac{24/25}{26/25}ight) = \cos^{-1}\left(\frac{12}{13}ight)$.હવે,પદાવલિ $\cos^{-1}\left(\frac{15}{17}ight) + \cos^{-1}\left(\frac{12}{13}ight)$ બને છે.સૂત્ર $\cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}ight)$ નો ઉપયોગ કરતા:$\cos^{-1}\left(\frac{15}{17} 	imes \frac{12}{13} - \sqrt{1 - \left(\frac{15}{17}ight)^2} \sqrt{1 - \left(\frac{12}{13}ight)^2}ight)$$= \cos^{-1}\left(\frac{180}{221} - \sqrt{1 - \frac{225}{289}} \sqrt{1 - \frac{144}{169}}ight)$$= \cos^{-1}\left(\frac{180}{221} - \sqrt{\frac{64}{289}} \sqrt{\frac{25}{169}}ight)$$= \cos^{-1}\left(\frac{180}{221} - \frac{8}{17} 	imes \frac{5}{13}ight)$$= \cos^{-1}\left(\frac{180}{221} - \frac{40}{221}ight) = \cos^{-1}\left(\frac{140}{221}ight)$.આમ,$\cos^{-1}\left(\frac{140}{221}ight)$ વિકલ્પોમાં આપેલ નથી,તેથી સાચો વિકલ્પ $(d)$ છે.

$\cos^{-1}\left(\frac{15}{17}\right) + 2\tan^{-1}\left(\frac{1}{5}\right) = $

Similar Questions

$\begin{aligned} & \text{જો } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & b>a, \text{ હોય તો } x= \end{aligned}$

જો $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$ હોય,તો $p^2 + q^2 + r^2 + 2pqr = $

જો $\sin^{-1}(1 - x) - 2\sin^{-1}x = \pi/2$ હોય,તો $x$ ની કિંમત શોધો:

જો $f(x) = 2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,જ્યાં $x > 1$,તો $f(5)$ ની કિંમત શોધો:

જો $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,હોય તો $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ ની કિંમત $........$ થાય.

Vedclass Test Series

Exam Paper Generator

Online Exam Module