Explanation
$$\textbf{Step 1 :- Formulating the information from the given question. }$$
$$\text{Suppose the third angle is p.}$$
$$\text{So from definition we know that the sum of all total three angles of a triangle is}$$ $$\pi^c$$.
$${\text{As 2}} \times {\text{3 > 1 So we can write }} {\tan^{ - 1}}2 + {\tan^{ - 1}}3 = \pi + {\tan^{ - 1}}\left( {\dfrac{{2 + 3}}{{1 - 2 \times 3}}} \right)$$
$$\mathbf{\left [ \tan^{-1} X + \tan^{-1}Y = \tan^{-1}\dfrac{X+Y}{1-(X\times Y)} \right ]}$$
$$ = \pi + {\tan^{ - 1}}\left( { - 1} \right)$$
$$ = \pi - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4}$$
$$\textbf{Step 2 :- Using formula of sum three angles of a triangle.}$$
$${\text{So p + }}\dfrac{{3\pi }}{4} = \pi $$
$$ \Rightarrow {\text{p = }}\pi {\text{ - }}\dfrac{{3\pi }}{4} = \dfrac{\pi }{4}$$
$$\textbf{Hence, The correct option is A}$$.
The value of $$\cot^{-1} \left[ \dfrac{\sqrt{1 - \sin x} +\sqrt{1 + \sin x}}{\sqrt{(1 - \sin x)} - \sqrt{(1 + \sin x)}} \right]$$ is
$$\\ y=sec(tan^{-1}x)\\\therefore sec(tan^{-1}x)tan(tan^{-1}x)\cdot (\frac{1}{1+x^2})\\ put x=1\\ then(\frac{dy}{dx})=sec(tan^{-1}1)tan(tan^{-1}1)\cdot (\frac{1}{1+1^2})\\=sec(\frac{\pi}{4})\cdot tan(\frac{\pi}{4})\cdot (\frac{1}{2})\\= \sqrt 2\cdot 1 \cdot (\frac{1}{2})=(\frac{1}{\sqrt2})$$
$${\sin ^{ - 1}}\left( {{4 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) + {\sin ^{ - 1}}\left( {{{16} \over {65}}} \right) = {\pi \over 2}$$
$${\sin ^{ - 1}}A + {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^{.2}}} + B\sqrt {1 - {B^2}} } \right)$$
$${\sin ^{ - 1}}\left( {{4 \over 5}\sqrt {1 - {{\left( {{5 \over {12}}} \right)}^2}} + {5 \over {13}}\sqrt {1 - {{\left( {{4 \over 5}} \right)}^2}} } \right)$$
$${\sin ^{ - 1}}\left( {{{63} \over {65}}} \right)$$
$${\sin ^{ - 1}}\left( {{{63} \over {65}}} \right) + {\sin ^{ - 1}}{{16} \over {65}}$$
Applying some formula
$${\sin ^{ - 1}}\left( {{{63} \over {65}}\sqrt {1 - {{\left( { - {{16} \over {65}}} \right)}^2}} + {{16} \over {65}}\sqrt {1 - {{\left( {{{63} \over {65}}} \right)}^2}} } \right)$$
$${\sin ^{ - 1}}(1)$$
$${\sin ^{ - 1}}\left( {\sin \pi /2} \right)$$
$$\pi /2$$
We have,
$${{\cos }^{-1}}\dfrac{2}{3}+{{\tan }^{-1}}\dfrac{1}{7}$$
We know that,
$${{\tan }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$$
Therefore,
$$ ={{\cos }^{-1}}\dfrac{2}{3}+{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{1+{{\left( \dfrac{1}{7} \right)}^{2}}}} \right) $$
$$ ={{\cos }^{-1}}\dfrac{2}{3}+{{\cos }^{-1}}\left( \dfrac{7}{\sqrt{50}} \right) $$
$${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left\{ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right\}$$
$$ ={{\cos }^{-1}}\left\{ \dfrac{2}{3}\times \dfrac{7}{\sqrt{50}}-\sqrt{1-{{\left( \dfrac{2}{3} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{7}{\sqrt{50}} \right)}^{2}}} \right\} $$
$$ ={{\cos }^{-1}}\left\{ \dfrac{14}{3\sqrt{50}}-\sqrt{\dfrac{5}{9}}\sqrt{\dfrac{1}{50}} \right\} $$
$$ ={{\cos }^{-1}}\left\{ \dfrac{14}{3\sqrt{50}}-\dfrac{\sqrt{5}}{3\sqrt{50}} \right\} $$
$$ ={{\cos }^{-1}}\left\{ \dfrac{14-\sqrt{5}}{3\sqrt{50}} \right\} $$
Hence, this is the answer.
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