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  • The value of the expression tan 1 divided 2 cos inverse 2 divided square root 5 is in to bracket Hint: tan theta divided 2 equal to square root of 1 minus cos theta divided 1 plus cos theta A. 2 plus square root 5 B. square root 5 minus 2

The value of the expression \tan \frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}} is \left [ Hint: \tan\frac{\theta}{2} =\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right ]

A. 2+ \sqrt{5}

B.\sqrt{5}-2

C.\frac{2+\sqrt{5}}{2}

D. \sqrt{5}+2

Answers (1)

Answer:(B)

We need to find , \tan \frac{1}{2}\cos^{-1}\frac{2}{\sqrt{5}}

Let, \cos^{-1}\frac{2}{\sqrt{5}}=A

\Rightarrow \cos A=\frac{2}{\sqrt{5}}

Also we need to find \tan\frac{A}{2}

We know that \tan\frac{\theta}{2}=\sqrt{\frac{\left ( 1-\cos \theta \right )}{1+\cos \theta}}

so, \tan^{-1}\frac{A}{2}=\sqrt{\frac{\left ( 1-\cos A \right )}{1+\cos A}}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\frac{\sqrt{5}-2}{\sqrt{5}}}{\frac{\sqrt{5}+2}{\sqrt{5}}}}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}

on rationalizing, 

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\left (\sqrt{5}-2 \right )\left ( \sqrt{5}+2 \right )}{\left (\sqrt{5}+2 \right )\left ( \sqrt{5}+2 \right )}}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{\left (\sqrt{5} \right )^{2}-2^{2}}{\left (\sqrt{5} ^{2}+2^{2}\right )}}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{5-4}{\left (\sqrt{5} ^{2}+2^{2}\right )}}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{\frac{1}{\left (\sqrt{5} ^{2}+2^{2}\right )}}

\Rightarrow \tan^{-1} \frac{A}{2}=\frac{1}{\sqrt{5} +2}

Again rationalizing

\Rightarrow \tan^{-1} \frac{A}{2}=\frac{1\left ( \sqrt{5}-1 \right )}{\left (\sqrt{5} +2 \right )\left ( \sqrt{5}-2 \right )}

\Rightarrow \tan^{-1} \frac{A}{2}=\frac{\left ( \sqrt{5}-2 \right)}{\left (\sqrt{5}^{2} -2^{2} \right )}

\Rightarrow \tan^{-1} \frac{A}{2}=\frac{\left ( \sqrt{5}-2 \right)}{\left (5-4 \right )}

\Rightarrow \tan^{-1} \frac{A}{2}=\sqrt{5}-2

 

 

 

 

 

Posted by

infoexpert24

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