Math, asked by IsabelGlickman, 1 month ago

Solve for θin radians 7 tan^2(θ) = 7

Answers

Answered by devichandru65
2

Step-by-step explanation:

The angles which lie between 0° and 90° are said to lie in the first quadrant. The angles between 90° and 180° are in the second quadrant, angles between 180° and 270° are in the third quadrant and angles between 270° and 360° are in the fourth quadrant: ... In the second quadrant, the values for sin are positive only.

Answered by Aryan0123
17

In this Question,

  • 7 tan²θ = 7

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Dividing by 7 on both sides,

→ tan²θ = 1

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Taking Square root on both sides;

→ √(tan²θ) = 1

→ tan θ = √1

→ tan θ = 1

→ θ = tan inverse (1)

θ = 45°

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∴ The value of θ is 45°

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ADDITIONAL INFORMATION:

\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \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}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

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