Question

If sinA+cosA = √2 sin(90–A), then obtain the value of cotA.

Answer 1

HELLO FRIEND ,

HERE IS YOUR ANSWER,

we can write, sin(90°-A) = cosA

therefore,

SinA + cosA = 2^(1/2)cosA
SinA. = [2^(1/2)-1]cosA
cotA. = root2 + 1

I HOPE YOU WILL UNDERSTAND .......!!!!!!!!!

Answer 2

$\sin \: A + \cos A = \sqrt{2} \sin(90 - A) \\ \\ \\ = > \sin \: A + \cos \: A = \sqrt{2} \cos \: A \\ \\ = > \frac{ \sin \: A + \cos \: A }{ \cos A} = \sqrt{2} \\ \\ \\ = > \frac{ \sin \: A }{ \cos \: A } + 1 = \sqrt{2} \\ \\ \\ = > \tan \: A = \sqrt{2} - 1$


Hence,


$\cot \: A = \frac{1}{ \tan \: A } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (as \: we \: know \: ) \\ \\ \\ = > \cot \: A = \frac{1}{ \sqrt{2} - 1 } \\ \\ \\ = > \cot \: A = \frac{ \sqrt{2 } + 1 }{( \sqrt{2} - 1)( \sqrt{2} + 1) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (by \: rationalazing) \\ \\ \\ = > \cot \: A = \frac{ \sqrt{2} + 1}{2 - 1} \\ \\ \\ = > \cot \: A = \sqrt{2} + 1$