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What is the value of 

\frac{\cos(A+B)}{\cos A \cos B}+\frac{\cos(B+C)}{\cos B \cos C}+\frac{\cos(C+A)}{\cos C \cos A} 

Option: 1

3-(\tan A \tan B+\tan B \tan C+\tan A \tan C)


Option: 2

3-(\cot A \cot B+\cot B \cot C+\cot A \cot C)


Option: 3

0


Option: 4

3


Answers (1)

best_answer

Trigonometric Ratio for Compound Angles (Part 1)

cos (α+ β) = cos α cos β - sin α sin β

 

Now,

\frac{\cos(A+B)}{\cos A \cos B}+\frac{\cos(B+C)}{\cos B \cos C}+\frac{\cos(C+A)}{\cos C \cos A}\\\\ First\ term\ of\ the\ above\ equation\\ \\\frac{\cos(A+B)}{\cos A \cos B}=\frac{\cos(A)\cos(B)-\sin A \sin B}{\cos A \cos B}\\\\ =1-\tan A \tan B\\\\ Similarly\ second\ term\ = 1-\tan B \tan C\\ \\and\ third\ term= 1-\tan A \tan C\\\\ \frac{\cos(A+B)}{\cos A \cos B}+\frac{\cos(B+C)}{\cos B \cos C}+\frac{\cos(C+A)}{\cos C \cos A}\\\\=1-\tan A \tan B+1-\tan B \tan C+1-\tan A \tan C\\\\ =3-\tan A \tan B-\tan B \tan C-\tan A \tan C\\\\ =3-(\tan A \tan B+\tan B \tan C+\tan A \tan C)

Posted by

shivangi.shekhar

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