Question

SecA tanB+tanA secB=2,secA secB+tanA tanB=k then k square=

Answer 1

$\textbf{Given:}$

$secA\,tanB+tanA\,secB=2$

$secA\,secB+tanA\,tanB=k$

$\textbf{To find:}$

$\text{The value of k^2 }$

$\textbf{Solution:}$

$\text{Consider,}$

$k^2$

$=(secA\,secB+tanA\,tanB)^2$

$=sec^2A\,sec^2B+tan^2A\,tan^2B+2\,secA\,secB\,tanA\,tanB$

$=sec^2A(1+tan^2B)+tan^2A(sec^2B-1)+2\,secA\,secB\,tanA\,tanB$

$=sec^2A+sec^2A\,tan^2B+tan^2A\,sec^2B-tan^2A+2\,secA\,secB\,tanA\,tanB$

$=(sec^2A-tan^2A)+(sec^2A\,tan^2B+tan^2A\,sec^2B+2\,secA\,tanB\,tanA\,secB)$

$=(sec^2A-tan^2A)+(secA\,tanB+tanA\,secB)^2$

$=(1)+(2)^2$

$=1+4$

$=5$

$\textbf{Answer:}$

$\boxed{\textbf{The value of \bf\,k^2 is 5}}$