Question

please solve this question...
And don't sent the wastage answer...
please help me for the correct answer....

Answer 1

Open it...
Go through it only ... Because except this, no other chances are here.....
Hope it will help You

Answer 2

$Given : tanA = \sqrt{2} - 1$

Now,

$= \ \textgreater \ \frac{1}{tanA} = \frac{1}{ \sqrt{2} - 1 } * \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1 }$

$= \ \textgreater \ \frac{ \sqrt{2} + 1 }{( \sqrt{2})^2 - (1)^2 }$

$= \ \textgreater \ \sqrt{2} + 1$


Now,

$= \ \textgreater \ tanA + \frac{1}{tanA} = \sqrt{2} - 1 + \sqrt{2} + 1$

$= \ \textgreater \ \sqrt{2} + \sqrt{2}$

$= \ \textgreater \ 2 \sqrt{2}$


Now,

$= \ \textgreater \ tanA + \frac{1}{tanA} = 2 \sqrt{2}$

$= \ \textgreater \ \frac{tan^2A + 1}{tanA} = 2 \sqrt{2}$

$= \ \textgreater \ \frac{sec^2A}{tanA} = 2 \sqrt{2}$

$= \ \textgreater \ \frac{1}{cos^2A} * \frac{cosA}{sinA} = 2 \sqrt{2}$

$= \ \textgreater \ \frac{1}{cosAsinA} = 2 \sqrt{2}$

$= \ \textgreater \ cosAsinA = \frac{1}{2 \sqrt{2} }$



Hope this helps!