Question

plz help in any way you can​

Answer 1

Given :-

$Cos \theta - Sin\theta = \sqrt{2}Sin\theta$

To prove :-

$Cos \theta + Sin \theta = \sqrt{2} Cos \theta$

Solution:-

→$Cos \theta - Sin\theta = \sqrt{2}Sin\theta$

→$Cos \theta = \sqrt{2}Sin\theta + Sin\theta$

→$Cos \theta = Sin \theta( \sqrt{2}+1)$

→$Cos \theta = Sin\theta \left(\sqrt{2}+1 \times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}\right)$

→$Cos \theta = Sin\theta \left(\dfrac{2-1}{\sqrt{2}-1}\right)$

→$Cos \theta = \dfrac{Sin\theta}{\sqrt{2}-1}$

→$Sin\theta = Cos \theta (\sqrt{2}-1)$

• Put the value of $Cos \theta$

→$Cos \theta + Sin \theta = \sqrt{2} Cos \theta$

→$Cos \theta + \sqrt{2}Cos \theta - Cos \theta = \sqrt{2}Cos \theta$

→$\sqrt{2}Cos \theta = \sqrt{2}Cos \theta$

hence,

proved....