Question

Pls solve the 16 th question it's urgent thank you!!!

Answer 1

$(tanA+\frac{1}{cosA} )^2+(tanA-\frac{1}{cosA} )^2=2(\frac{1+sin^2A}{1-sin^2A})$

solving L.H.S

=$(tanA+\frac{1}{cosA} )^2+(tanA-\frac{1}{cosA} )^2</p><p>=[tex](\frac{sinA}{cosA}+ \frac{1}{cosA})^2+(\frac{sinA}{cosA}- \frac{1}{cosA})^2$    [∵tanA=$\frac{sinA}{cosA}$]

=$(\frac{sinA+1}{cosA} )^2+(\frac{sinA-1}{cosA} )^2$

=$\frac{(sinA+1)^2}{cos^2A} + \frac{(sinA-1)^2}{cos^2A}$

=$\frac{sin^2A+1+2sinA}{cos^2A} +\frac{sin^2A+1-2sinA}{cos^2A}$

     [∵$(a^2+b^2)=a^2+b^2+2ab$]   [∵$(a^2-b^2)=a^2+b^2-2ab$]

=$\frac{sin^2A+1+2sinA+sin^2A+1-2sinA}{cos^2A}$

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

=$2(\frac{1+sin^2A}{1-sin^2A})$   [∵$sin^2A+cos^2A=1$]

HOPE IT WILL HELP YOU