Question

If $A+B+C={90}^0,then\sum \frac{\cos (A+B)}{\cos A\cos B}$ is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Given $A+B+C={90}^o$

$\sum \frac{\cos (A+B)}{\cos A\cos B}$

$=\frac{\cos (A+B)}{\cos A.\cos B}+\frac{\cos (B+C)}{\cos B.\cos C}+\frac{\cos (C+A)}{\cos C.\cos A}$

$=\frac{\cos A\cos B-\sin A\sin B}{\cos A.\cos B}+\frac{\cos B.\cos C-\sin B\sin C}{\cos B.\cos C}+\frac{\cos C.\cos A-\sin C.\sin A}{\cos C.\cos A}$

$=1-\frac{\sin A.\sin B}{\cos A.\cos B}+1-\frac{\sin B.\sin C}{\cos B.\cos C}+1-\frac{\sin C.\sin A}{\cos C.\cos A}$

$=3-\frac{\sin A}{\cos A}.\frac{\sin B}{\cos B}-\frac{\sin B}{\cos C}.\frac{\sin C}{\cos C}-\frac{\sin C}{\cos C}.\frac{\sin A}{\cos A}$$

$=3-\tan A.\tan B-\tan B.\tan C-\tan C.\tan A$

$=3-[\tan A.\tan B+\tan B.\tan C+\tan C\tan A].......\left(I\right)$

$A+B+C={90}^o$

$A+B={90}^o-C$

Apply tan on both sides

$\tan (A+B)=\tan ({90}^o-C)$

$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=\cot C$

$\Rightarrow \frac{\tan A+\tan B}{1-\tan A.\tan B}=\frac{1}{\tan C}$

$\Rightarrow \tan A.\tan C+\tan B.\tan C=1-\tan A.\tan B$

$\Rightarrow \tan A.\tan B+\tan A.\tan C+\tan B.\tan C=1.......\left(II\right)$

Substitute equation $\left(II\right)$ in $\left(I\right)$

$\Rightarrow \sum \frac{\cos (A+B)}{\cos A.\cos B}=3-[\tan A.\tan B+\tan B\tan C+\tan C.\tan A]$

$=3-1$

$=2$

$\therefore \sum \frac{\cos (A+B)}{\cos A.\cos B}=2$