Question

If cos theta+ Sin theta = √2 cos theta , Show that Cos theta - Sin theta = √2sin theta.​

Answer 1

Solution:-

It is given that

Cosθ + Sinθ = √2 Cosθ

Squaring both side we get

→ (Cosθ + Sinθ)² = 2 Cos²θ

→ Cos²θ + Sin²θ + 2 Cosθ Sinθ = 2 Cos²θ

→ Cos²θ - 2 Cosθ Sinθ = Sin²θ

→Cos² θ - 2Cosθ Sinθ + Sin²θ = 2Sin²θ

→( Cosθ - Sinθ)² = 2Sin²θ

→ Cosθ - Sinθ = √2 Sinθ

Additional Information !!

Trigonometry Ratios

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$