Question

If sine + caso = √2cos (90° - 0), find cote.​

Answer 1

\begin{gathered} \sin( \alpha ) + \cos( \alpha ) = \sqrt{2} \cos( 90 - \alpha ) \\ \\ \sin( \alpha ) + \cos( \alpha ) = \sqrt{2} \sin( \alpha ) \\ \\ \cos( \alpha ) = \sqrt{2} \sin( \alpha ) - \sin( \alpha ) \\ \\ \cos( \alpha ) = \sin \alpha ( \sqrt{2} - 1) \\ \\ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } = ( \sqrt{2} - 1) \\ \\ \cot( \alpha ) = ( \sqrt{2} - 1)\end{gathered}sin(α)+cos(α)=2cos(90−α)sin(α)+cos(α)=2sin(α)cos(α)=2sin(α)−sin(α)cos(α)=sinα(2−1)sin(α)cos(α)=(2−1)cot(α)=(2−1)

Answer 2

Answer :

cotθ = √2 - 1

Solution :

• Given : sinθ + cosθ = √2cos(90°- θ)
• To find : cotθ = 0

We have ,

=> sinθ + cosθ = √2cos(90°- θ)

=> sinθ + cosθ = √2sinθ

=> cosθ = √2sinθ - sinθ

=> cosθ = (√2 - 1)sinθ

=> cosθ/sinθ = √2 - 1

=> cotθ = √2 - 1

Hence , cotθ = √2 - 1