Question

if √3tan. Tita = 1 ,find the value of sin3 Tita + cos 2 Tita​

Answer 1

Answer:

$\sf The \ value \ of \ 3sin \theta \ + \ 2cos \theta \ = \ 3 \ + \ 2 \dfrac{\sqrt 3}{2}$

Correct Question :

$\sf If \ \sqrt{3} tan \theta \ = \ 1, \ find \ the \ value \ of \ 3sin \theta \ + \ 2 cos \theta .$

Solution :

According the the question,

We know that, $\sf tan \theta \ = \ \dfrac {Perpendicular}{Base}$

So,

$\implies \sf tan \theta \ = \ \dfrac {1}{\sqrt 3}$

First, we need to find the value of hypotenuse = ?

Using Pythagoras theorem,

$\sf (Hypotenuse)^2 \ = \ (Base)^2 + (Perpendicular)^2$

$\sf (Hypotenuse)^2 \ = \ (\sqrt 3)^2 \ + \ (1)^2$

$\sf (Hypotenuse)^2 \ = \ \sqrt{3} \times \sqrt{3} \times 1$

$\sf (Hypotenuse)^2 \ = \ 3 \ + \ 1$

$\sf (Hypotenuse)^2 \ = \ 4$

$\sf Hypotenuse \ = \ \sqrt{4}$

$\large {\boxed {\bf Hypotenuse \ = \ 2}}$

Now,

$\sf To \ find \ the \ value \ of \ 3sin \ + \ 2cos \ = \ ?$

$\implies \sf 3 sin \theta \ + \ 2 cos \theta$

$\implies \bf 3 \times \dfrac {perpendicular}{Hypotenuse} \ + \ 2 \times \dfrac {base}{Hypotenuse}$

$\implies \sf 3 \times \dfrac {1}{2} \ + \ {\cancel {2}} \times \dfrac {\sqrt 3}{\cancel 2}$

$\implies \sf \dfrac {3}{2} \times \sqrt{3}$

$\implies \sf 3 \ + \ 2 \ \dfrac {\sqrt 3}{2}$

$\large {\boxed {\bf 3 \ + \ 2 \ \dfrac {\sqrt 3}{2}}}$

Answer 2

Solution :

Given :

If √3 tan Ѳ = 1

Find :

3sin Ѳ + 2cos Ѳ

Explanation :

As we know that tan Ѳ = Perpendicular/base

A/q

• tan Ѳ = 1/√3

Using by Pythagoras Theorem :

(Hypotenuse)² = (Base)² + (Perpendicular)²

(Hypotenuse)² = (√3)² + (1)²

(Hypotenuse)² = √3 × √3 + 1

(Hypotenuse)² = 3 + 1

(Hypotenuse)² = 4

Hypotenuse = √4

Hypotenuse = 2

Now,

3sin Ѳ + 2cos Ѳ

3 × Perpendicular/hypotenuse + 2 × base/hypotenuse

3 × 1/2 + 2 × √3/2

3/2 + √3

3 + 2√3/2