Question

If $A+B={90}^o$ then prove that
$\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}}=\sec A$

Answer 1

Given $A+B=90$ $\Rightarrow$ $B=90-A$
LHS$=\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}}$
$=\sqrt{\frac{\tan A.\tan (90-A)+\tan A.\cot (90-A)}{\sin A.\sec (90-A)}}$

$=\sqrt{\frac{\tan A\cdot \cot A+\tan A\cdot \tan A}{\sin A\text{cosec}A}}$

$=\sqrt{\frac{\tan A.\frac{1}{\tan A}+{\tan }^{2}A}{\sin A.\frac{1}{\sin A}}}$

$=\sqrt{\frac{1+{\tan }^{2}A}{1}}$
$=\sqrt{{\sec }^{2}A}=\sec A=$ RHS

Hence, $=\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}}=\sec A$