Question

Prove : $(sinA+cosecA{)}^2+(cosA+secA{)}^2=7+ta{n}^{2}A+co{t}^{2}A$

Answer 1

$(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$

$LHS:{(\sin A+\frac{1}{\sin A})}^2+{(\cos A+\frac{1}{\cos A})}^2[\because \csc A=\frac{1}{\sin A},\sec A=\frac{1}{\cos A}]$

$=\frac{({\sin }^{2}A+1{)}^2}{{\sin }^{2}A}+\frac{(1+{\cos }^{2}A)}{{\cos }^{2}A}=\frac{{\cos }^{2}A(1+{\sin }^{2}A)+{\sin }^{2}A(1+{\cos }^{2}A{)}^2}{{\sin }^{2}A.{\cos }^{2}A}$

$=\frac{{\sin }^{2}A+{\sin }^{4}A.{\cos }^{2}A+2{\sin }^{2}A.{\cos }^{2}A+{\sin }^{2}A+{\cos }^{4}A.{\sin }^{2}A+2{\sin }^{2}A{\cos }^{2}A}{{\sin }^{2}A.{\cos }^{2}A}$

$=\frac{({\cos }^{2}A+{\sin }^{2}A)+4{\sin }^{2}A.{\cos }^{2}A+{\sin }^{2}A.{\cos }^{2}A({\sin }^{2}A+{\cos }^{2}A)}{{\sin }^{2}A.{\cos }^{2}A}$

$=\frac{1+4{\sin }^{2}A.{\cos }^{2}A+{\sin }^{2}A.{\cos }^{2}A}{{\sin }^{2}A.{\cos }^{2}A}=\frac{1+5{\sin }^{2}A{\cos }^{2}A}{{\sin }^{2}A.{\cos }^{2}A}=LHS$

$RHS:7+{\tan }^{2}A+{\cot }^{2}A=7+\frac{{\sin }^{2}A}{{\cos }^{2}A}+\frac{{\cos }^{2}A}{{\sin }^{2}A}$

$=\frac{7{\sin }^{2}A.{\cos }^{2}A+{\sin }^{4}A+{\cos }^{4}A}{{\cos }^{2}A.{\sin }^{2}A}=\frac{5{\sin }^{2}A.{\cos }^{2}A+({\sin }^{2}A+{\cos }^{2}A{)}^2}{{\sin }^{2}A.{\cos }^{2}A}$

$=\frac{1+5{\sin }^{2}A.{\cos }^{2}A}{{\sin }^{2}A.{\cos }^{2}A}=R.H.S$

Hence $LHS=RHS$