If 3sec2θ + tanθ - 7 = 0, 0° < θ < 90°, then what is the value of  \((\frac{2 sinθ +3 cosθ} {cosecθ +secθ})\) ?

If 3sec2θ + tanθ - 7 = 0, 0° < θ < 90°, then what is the value of  \((\frac{2 sinθ +3 cosθ} {cosecθ +secθ})\) ? Correct Answer <span class="math-tex">\(\frac{5}{4}\)</span>

Given:

3sec2θ + tanθ - 7 = 0

Calculation:

3sec2θ + tanθ - 7 = 0

Put θ = 45°

⇒ 3 sec²45° + tan 45° - 7

= (3 × 2) + 1 - 7

= 6 + 1 - 7

= 0 

So, (2 sin θ + 3 cos θ)/(cosec θ + sec θ) = (2 sin 45° + 3 cos 45°)/(cosec 45° + sec 45°)

= /(√2 + √2)

= (2/√2 + 3/√2)/2√2

= (5/√2)/2√2

= 5/(√2 × 2√2)

= 5/4

∴ The value of (2 sin θ + 3 cos θ)/(cosec θ + sec θ) is 5/4