Question

ᴘʟᴇᴀsᴇ sᴏʟᴠᴇ ᴛʜɪs ǫᴜᴇsᴛɪᴏɴ :

cos²θ - 3cosθ + 2 / sin²θ = 1 ​

Answer 1

• GIVEN:-

$\large{\sf{\frac{cos²θ - 3cosθ + 2}{sin^2\theta}=1}}$

• To Find:-

Evaluation.

• SOLUTION:-

$\large\Longrightarrow{\sf{\frac{cos²θ - 3cosθ + 2}{sin^2\theta}=1}}$

$\large\Longrightarrow{\sf{\frac{cos²θ - 3cosθ + 2}{1-cos^2\theta}=1}}$

$\small\because{\sf{(sin^2\theta=1-cos^2\theta)}}$

By cross multiplication,

$\small\Longrightarrow{\sf{cos²θ – 3cosθ + 2= 1-cos^2\theta}}$

$\small\Longrightarrow{\sf{cos²θ – 3cosθ + 2-1+cos^2\theta=0}}$

$\small\Longrightarrow{\sf{2cos²θ – 3cosθ +1=0}}$

By splitting the middle term,

$\small\Longrightarrow{\sf{2cos²θ – 2cosθ – cosθ + 1 = 0}}$

$\small\Longrightarrow{\sf{2cosθ (cosθ – 1) – 1 (cosθ – 1) = 0}}$

$\small\Longrightarrow{\sf{(2 cosθ – 1) (cosθ – 1) = 0}}$

$\large\boxed{\sf{cosθ = \frac{1}{2}}}$

or,

$\large\boxed{\sf{cosθ = 1}}$

as, cosθ ≠ 1 (because θ > 0)

=> cosθ = cos 1/2

∴ cosθ = cos 60°

$\huge\red\therefore\boxed{\bf{\red{\theta=60\degree}}}$