Question

If $A+B={90}^o$, prove that

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

Answer 1

$A+B={90}^o$

$B={90}^o-A$

LHS :

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

$=\sqrt{\frac{tanAtan({90}^o-A)+tanAcot({90}^o-A)}{sinAsec({90}^o-A)}-\frac{si{n}^2({90}^o-A)}{co{s}^{2}A}}$

$=\sqrt{\frac{tanAcotA+tanAtanA}{sinAcosecA}-\frac{co{s}^{2}A}{co{s}^{2}A}}$

$=\sqrt{1+ta{n}^{2}A-1}$

$=\sqrt{ta{n}^{2}A}=tanA$