Question

Cos37°+sin37°/Cos37°-sin37° =cot8°

Answer 1

$LHS = \frac{cos 37\degree + sin 37 \degree }{cos 37\degree - sin 37 \degree} \\= \frac{cos 37\degree + sin (90 - 53) \degree }{cos 37\degree - sin (90- 53) \degree} \\= \frac{cos 37\degree + cos 53 \degree }{cos 37\degree - cos 53 \degree} \\= \frac{ 2 cos \Big( \frac{37+53}{2}\Big) cos \Big( \frac{37-53}{2}\Big) }{ -2 sin \Big( \frac{37+53}{2}\Big) sin \Big( \frac{37-53}{2}\Big) }\\= \frac{ 2 cos \Big( \frac{90}{2}\Big) cos \Big( \frac{-16}{2}\Big) }{ -2 sin \Big( \frac{90}{2}\Big) sin \Big( \frac{-16}{2}\Big) }\\= \frac{ cos 45 \degree cos (-8) \degree }{ - sin 45 \degree sin (-8) \degree } \\= \frac{\frac{1}{\sqrt{2}} \times cos 8 \degree }{ - \frac{1}{\sqrt{2}} \times ( - sin \: 8 \degree ) }\\= \frac{cos 8 \degree }{ sin 8 \degree } \\= cot 8 \degree \\ = RHS$

$Hence\: proved$

•••♪

Answer 2

Step-by-step explanation:

GIVEN,

$\frac{ \cos(37) + \sin(37) }{ \cos(37) - \sin(37) }$

$= \frac{ \cos(90 -53) + \sin(37) }{ \cos(90 - 53) - \sin(37) }$

$= \frac{ \sin(53) + \sin(37) }{ \sin(53) - \sin(37) }$

$= \frac{2 \sin( \frac{53 + 37}{2}) \cos( \frac{53 - 37}{2} ) }{2 \sin( \frac{53 - 37}{2} ) \cos( \frac{53 + 37}{2} ) }$

$= \frac{ \sin(45) \cos(8) }{ \sin(8) \cos(45) }$

$= \tan(45) \cot(8)$

$= 1 \times \cot(8)$

$= \cot(8)$

HOPE THAT HELPS!!