Question

Q. If cosA + sinA = √2 sin(90° - A), show that cosA - sinA = √2 cos(90° - A).

Answer 1

Answer:

$</p><p>\begin{gathered} \sin \: A + \cos A = \sqrt{2} \sin(90 - A) \\ \\ \\ = > \sin \: A + \cos \: A = \sqrt{2} \cos \: A \\ \\ = > \frac{ \sin \: A + \cos \: A }{ \cos A} = \sqrt{2} \\ \\ \\ = > \frac{ \sin \: A }{ \cos \: A } + 1 = \sqrt{2} \\ \\ \\ = > \tan \: A = \sqrt{2} - 1 \\ \\ \end{gathered}sinA+cosA=2sin(90−A)=>sinA+cosA=2cosA=>cosAsinA+cosA=2=>cosAsinA+1=2=>tanA=2−1</p><p></p><p></p><p>Hence,</p><p></p><p></p><p>\begin{gathered} \cot \: A = \frac{1}{ \tan \: A } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (as \: we \: know \: ) \\ \\ \\ = > \cot \: A = \frac{1}{ \sqrt{2} - 1 } \\ \\ \\ = > \cot \: A = \frac{ \sqrt{2 } + 1 }{( \sqrt{2} - 1)( \sqrt{2} + 1) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (by \: rationalazing) \\ \\ \\ = > \cot \: A = \frac{ \sqrt{2} + 1}{2 - 1} \\ \\ \\ = > \cot \: A = \sqrt{2} + 1\end{gathered}cotA=tanA1(asweknow)=>cotA=2−11=>cotA=(2−1)(2+1)2+1(byrationalazing)=>cotA=2−12+1=>cotA=2+1</p><p>$