If √2tan2θ = √6 and 0⁰ ≤ 2θ ≤ 90⁰, find the value of sin θ + √3cos θ – 2 tan2 θ.
Answers
Step-by-step explanation:
Hey mate,
We have,
We have,3
We have,3
We have,3 sinθ−cosθ=0
We have,3 sinθ−cosθ=03
We have,3 sinθ−cosθ=03
We have,3 sinθ−cosθ=03 sinθ=cosθ
We have,3 sinθ−cosθ=03 sinθ=cosθ3
We have,3 sinθ−cosθ=03 sinθ=cosθ3
We have,3 sinθ−cosθ=03 sinθ=cosθ3 =
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθ
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθ
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ=
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π θ=nπ+
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π θ=nπ+ 6
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π θ=nπ+ 6π
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π θ=nπ+ 6π
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π θ=nπ+ 6π where n ε Z
We have,3 sinθ−cosθ=03 sinθ=cosθ3 = sinθcosθ 3 =cotθtanθ= 3 1 tanθ=tan 6π θ=nπ+ 6π where n ε ZHence, this is the answer.