Find The Value Of Sinh Left Log Left 2+sqrt5 Right Class 11 Maths CBSE

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We then substitute $\log \left( 2+\sqrt{5} \right)$ in place of ‘y’ and then make use of the results $-\log a=\log \dfrac{1}{a}$ and ${{e}^{\log a}}=a$ to proceed through the problems. We then make use of the result ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and then perform the necessary calculations to get the required result.Complete step-by-step answer:According to the problem, we are asked to find the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$, where logarithm is to the natural base e.We know that the sine hyperbolic function is defined as $\sinh \left( y \right)=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}$. Let us use this definition to find the value of the given $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$.So, we have \[\sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{{{e}^{\log \left( 2+\sqrt{5} \right)}}-{{e}^{-\log \left( 2+\sqrt{5} \right)}}}{2}\] ---(1).We know that $-\log a=\log \dfrac{1}{a}$. Let us use this result in equation (1).\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{{{e}^{\log \left( 2+\sqrt{5} \right)}}-{{e}^{\log \left( \dfrac{1}{2+\sqrt{5}} \right)}}}{2}\] ---(2).We know that ${{e}^{\log a}}=a$. Let us use this result in equation (2).\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{2+\sqrt{5}-\left( \dfrac{1}{2+\sqrt{5}} \right)}{2}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( \dfrac{{{\left( 2+\sqrt{5} \right)}^{2}}-1}{2+\sqrt{5}} \right)}{2}\] ---(3).We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Let us use this result in equation (3).\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( \dfrac{{{2}^{2}}+{{\left( \sqrt{5} \right)}^{2}}+2\left( 2 \right)\left( \sqrt{5} \right)-1}{2+\sqrt{5}} \right)}{2}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( 4+5+4\sqrt{5} \right)-1}{2\left( 2+\sqrt{5} \right)}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{8+4\sqrt{5}}{4+2\sqrt{5}}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=2\].So, we have found the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$ as 2.∴ The correct option for the given problem is (a).So, the correct answer is “Option (a)”.Note: If the given logarithm is not to the natural base, then the result will not be the same as we get in this problem. We should always take logarithms to the natural base if nothing is mentioned in the problem about the base while solving problems related to hyperbolic functions. We can also solve this problem as shown below:Let us assume $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)=x$.$\Rightarrow \log \left( 2+\sqrt{5} \right)={{\sinh }^{-1}}\left( x \right)$.We know that ${{\sinh }^{-1}}\left( x \right)=\log \left( x+\sqrt{{{x}^{2}}+1} \right)$.$\Rightarrow \log \left( 2+\sqrt{5} \right)=\log \left( x+\sqrt{{{x}^{2}}+1} \right)$.$\Rightarrow 2+\sqrt{5}=x+\sqrt{{{x}^{2}}+1}$.Comparing rational and irrational parts on both sides we get,$\Rightarrow x=2$.Recently Updated PagesWhy is there a time difference of about 5 hours between class 10 social science CBSEIn cricket, what is a "pink ball" primarily used for?In cricket, what is the "new ball" phase?In cricket, what is a "death over"?What is the "Powerplay" in T20 cricket?In cricket, what is a "super over"?Why is there a time difference of about 5 hours between class 10 social science CBSEIn cricket, what is a "pink ball" primarily used for?In cricket, what is the "new ball" phase?In cricket, what is a "death over"?What is the "Powerplay" in T20 cricket?In cricket, what is a "super over"?

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