If tan A = 4 /3, tanB = 1/ 7,then A - B =

\frac{π}{4}

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\frac{π}{2}

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π

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\frac{2 π}{3}

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Solution

The correct option is C $\frac{π}{4}$
$tan (A - B) = \frac{tan A - tan B}{1 + tan A . tan B}$

$\Rightarrow$ $tan (A - B) = \frac{\frac{4}{3} - \frac{1}{7}}{1 + \frac{4}{3} \times \frac{1}{7}}$

$\Rightarrow$ $tan (A - B) = \frac{\frac{28 - 3}{21}}{1 + \frac{4}{21}}$

$\Rightarrow$ $tan (A - B) = \frac{\frac{25}{21}}{\frac{25}{21}}$

$\Rightarrow$ $tan (A - B) = 1$

$\Rightarrow$ $A - B = {tan}^{- 1} (1)$

$\Rightarrow$ $A - B = \frac{π}{4}$

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^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

A - B = \frac{π}{4}

(1 + tan A) (1 - tan B) =

π / 4

sin A = \frac{3}{5}, tan B = \frac{1}{2}

\frac{π}{2} < A < π

π < B < \frac{3 π}{2}

8 tan A - \sqrt{5} sec B =

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