Question 69

If (tanA + tanB)/(1 -­ tanAtanB) = x, then the value of x is

Solution

Expression : (tanA ­+ tanB)/(1 - tanAtanB) = x

= $$(\frac{sin A}{cos A} + \frac{sin B}{cos B}) \div (1 - \frac{sin A sin B}{cos A cos B})$$

= $$(\frac{sin Acos B + cos Asin B}{cos Acos B}) \div (\frac{cos Acos B - sin Asin B}{cos Acos B})$$

= $$[\frac{sin(A + B)}{cos A cos B}] \div [\frac{cos(A + B)}{cos A cos B}]$$

= $$[\frac{sin(A + B)}{cos A cos B}] \times [\frac{cos A cos B}{cos(A + B)}]$$

= $$\frac{sin(A + B)}{cos(A + B)} = tan(A + B)$$

=> Ans - (A)

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