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  • Find the value of <img alt="\cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)=" src="https://learn.careers360.com/latex-image/?%5Ccos%20%5E%7B-1%7D%5Cleft%28%5Cfrac%7B1

Find the value of \cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)=

Option: 1

{\frac{\pi}{2}} \\


Option: 2

{\cos ^{-1}\left(\frac{171}{221}\right)}


Option: 3

\\ {\frac{\pi}{4}}


Option: 4

None of these


Answers (1)

best_answer

 

 

Sum and difference of angles in terms of arccos -

Sum and difference of angles in terms of arccos

 

\\1.\;\;\cos ^{-1} x+\cos ^{-1} y=\left\{\begin{array}{ll}{\cos ^{-1}\{x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\}} & {\text { if }-1 \leq x, y \leq 1 \text { and } x+y \geq 0} \\\\ {2 \pi-\cos ^{-1}\{x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\}} & {\text { if }-1 \leq x, y \leq 1 \text { and } x+y \leq 0}\end{array}\right.\\\\\\\\2.\;\; \cos ^{-1} x-\cos ^{-1} y=\left\{\begin{array}{ll}{\cos ^{-1}\{x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\}} & {\text { if }-1 \leq x, y \leq 1 \text { and } x \leq y}\\ \\ {-\cos ^{-1}\{x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\}} & {\text { if }-1 \leq y \leq 0,0<x \leq 1 \text { and } x \geq y}\end{array}\right.

 

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{\cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)=\cos ^{-1}\left(\frac{15}{17}\right)+\cos ^{-1}\left(\frac{1-1 / 25}{1+1 / 25}\right)} \\ \because \cos ^{-1} x+\cos ^{-1} y=\left\{\begin{array}{ll}{\cos ^{-1}\{x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\}} & {\text { if }-1 \leq x, y \leq 1 \text { and } x+y \geq 0} \\ {2 \pi-\cos ^{-1}\{x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\}} & {\text { if }-1 \leq x, y \leq 1 \text { and } x+y \leq 0}\end{array}\right.\\ {=\cos ^{-1}\left(\frac{15}{17}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\\ =\cos ^{-1}\left(\frac{15}{17} \times \frac{12}{13}-\sqrt{1-\left(\frac{15}{17}\right)^{2}} \sqrt{\left.1-\left(\frac{12}{13}\right)^{2}\right)}\right)} \\ {=\cos ^{-1}\left(\frac{140}{221}\right)}

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Deependra Verma

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