Question

If $A+B={90}^{\circ }$, prove that $\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{{\sin }^{2}B}{{\cos }^{2}A}}=\tan A$

Answer 1

$A=90-B$

$L.H.S$

$\Rightarrow tanA=cotB;tanB=cotA,\cos A=\sin B,\sin A=\cos B$

$\Rightarrow \sqrt{\frac{tanAtanB+tanAtanB}{sinAsecB}-\frac{si{n}^{2}B}{co{s}^{2}A}}$

$=\sqrt{\frac{ta{n}^{2}A+1}{sinAcosecA}-\frac{si{n}^{2}B}{si{n}^{2}B}}$

$=\sqrt{{\sec }^2-1}=\sqrt{{\tan }^{2}A}$

$=tanA=R.H.S$

$\therefore$ $\Rightarrow \sqrt{\frac{tanAtanB+tanAtanB}{sinAsecB}-\frac{si{n}^{2}B}{co{s}^{2}A}}=\tan A$

Hence Proved