Question

0° <
$\alpha$
< 90° and
$\sqrt{2}$
tan 2
$\alpha$
=
$\sqrt{6}$
then find the value of sin
$\sin\alpha$
$+ \sqrt{3} \cos( \alpha )$
​

Answer 1

Answer:

Sin θ + √3(Cos θ) = 2

Step-by-step explanation:

We have,

√2 Tan 2θ = √6

Now,

√2 Tan 2θ = √6

Tan 2θ = √6/√2

Tan 2θ = √(6/2)

Tan 2θ = √3

But we know that,

Tan 60° = √3

So,

Tan 2θ = Tan 60°

Since they are equal now,

2θ = 60°

θ = 60°/2

θ = 30°

Now, we need to find,

Sin θ + √3(Cos θ)

= Sin 30° + √3(Cos 30°)

= (1/2) + √3[(√3)/2]

= (1/2) + (3/2)

= (1 + 3)/2

= 4/2

= 2

Hence,

Sin θ + √3(Cos θ) = 2

Hope it helped and believing you understood it........All the best