Question

If sin¢ + cos¢ = √2sin (90° - ¢) show that cot¢ = √2 + 1

Answer 1

Hiii friend !!!



Sin theta + Cos theta = ✓2 Sin (90-theta)



Sin theta + Cos theta = ✓2 Cos theta



Dividing both sides by Sin theta, we get ;


=> Sin theta / Sin theta + Cos theta /Sin theta = ✓2 Cos theta / Sin theta



=> 1 + Cot theta = ✓2 Cot theta [ Sin¢/Cos¢= Cot



=> (✓2-1) Cot theta = 1


=> Cot theta = 1/✓2-1



Rationalizing the denominator ✓2-1 we get,


=> Cot theta = 1/✓2-1×✓2+1/✓2+1


=> Cot theta = (✓2+1)/(✓2-1)(✓2+1)




=> Cot theta = (✓2+1)/(✓2)²-(1)²



=> Cot theta = (✓2+1)......PROVED.....




★ HOPE IT WILL HELP YOU ★

Answer 2

sin¢ + cos¢ = √2 sin(90 - ¢)

We know,
$\sin(90 - \alpha ) = \cos( \alpha )$

sin¢ + cos¢ = √2 cos¢

(sin¢ + cos¢)/cos¢ = √2

sin¢/cos¢ + cos¢/cos¢ = √2


We know,
$\frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \tan( \alpha )$

tan¢ + 1 = √2

tan¢ = √2 - 1


We know,
$\tan( \alpha ) = \frac{1}{ \cot( \alpha ) }$

$\frac{1}{ \tan( \alpha ) } = \frac{1}{ \sqrt{2} - 1 }$

By rationalization,

$\frac{1}{ \tan( \alpha ) } = \frac{ \sqrt{2 } + 1}{ {( \sqrt{2} )}^{2} - {1}^{2} } \\ \\ \\ \frac{1}{ \tan( \alpha ) } = \frac{ \sqrt{2} + 1}{2 - 1} \\ \\ \cot( \alpha ) = \frac{ \sqrt{2} + 1}{1} \\ \\ \\ \cot( \alpha ) = \sqrt{2} + 1$


Hence, proved.




I hope this will help you


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