Question
A triangle ABC is right angles at B; find the value of
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B}$
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B}$
✓
Answer
Since, $ABC$ is a right angled triangle, right angled at $B.$
So, $A + C = 90^\circ$
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B} $
$=\frac{\sec \left(90^{\circ}-C\right) \cdot \cos e c C-\tan \left(90^{\circ}-C\right) \cdot \cot C}{\sin 90^{\circ}}$
$ =\frac{\cos e c C \cdot \cos e c C-\cot C \cdot \cot C}{1} $
$ =1\left(\because \operatorname{cosec}^2 \theta-\cot ^2 \theta=1\right)$
So, $A + C = 90^\circ$
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B} $
$=\frac{\sec \left(90^{\circ}-C\right) \cdot \cos e c C-\tan \left(90^{\circ}-C\right) \cdot \cot C}{\sin 90^{\circ}}$
$ =\frac{\cos e c C \cdot \cos e c C-\cot C \cdot \cot C}{1} $
$ =1\left(\because \operatorname{cosec}^2 \theta-\cot ^2 \theta=1\right)$
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