Question

if 7 sin^2 0+3cos^2 0=4,then evaluate sec0+cosec0​

Answer 1

Required Answer:-

Given:

• 7sin²θ + 3cos²θ = 4

To Find:

• The value of sec θ + cosec θ.

Solution:

Given that,

⟹ 7sin²θ + 3cos²θ = 4

This can be written as,

⟹ 3sin²θ + 3sin²θ + 4sin²θ = 4

Now, we know that,

⟹ sin²θ + cos²θ = 1

So,

⟹ 3 × 1 + 4sin²θ = 4

⟹ 4sin²θ = 1

⟹ sin²θ = 1/4

⟹ sin θ = 1/2

From the Trigonometry Ratio Table,

⟹ sin θ = sin 30°

⟹ θ = 30°

Therefore,

sec θ + cosec θ

= sec 30° + cosec 30°

= 2/√3 + 2

= 2√3/3 + 2

= (6 + 2√3)/3

Hence,

⟹ sec θ + cosec θ = (6 + 2√3)/3

Answer:

• sec θ + cosec θ = (6 + 2√3)/3

Trigonometry Ratio Table:

$\sf Trigonometry\: Value \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$