Question

If sin theta + cos theta = √2cos(90° - theta) , cot theta​

Answer 1

$\tt \: \blue {GIVEN }\\ sin \theta + cos \theta = √2cos(90° - \theta) \\ \\ \tt \blue{ \: TO \: \: FIND} \\ \large \bf cot \: \theta \\ \\ \tt \: \blue{SOLUTION} \\ \bf \: sin \: \theta + cos \: \theta = √2cos \: (90 - \theta ) \\ \\ \pmb{ \boxed {cos(90° -\theta ) = sin\theta} }\\ \\ \bf \: sin \: \theta + cos \: \theta = √2sin \: \theta \\ \bf \: cos \theta = √2sin \theta - sin \theta \\ \bf \: cos \theta = sin \theta (√2 - 1) \\ \bf \: cos \: \theta / sin \: \theta = √2 - 1 \\ \bf \: cot \: \theta = √2 - 1 \\ \bf \: cot \: \theta = 1.41 - 1 \\ \bf \: cot \: \theta \: ≈ \: 0.41$

Answer 2

Step-by-step explanation:

sinθ+cosθ=√2cos(90−θ)

cos(90°−θ)=sin theta

• cos(90°−θ)=sinθ

sinθ+cosθ=√2sinθ

cosθ=√2sinθ−sinθ

cosθ=sinθ(√2−1)

cosθ/sinθ=√2−1

cotθ=√2−1

cotθ=1.41−1

cotθ≈0.41