Question

If cos thita =√3 upon 2 then find the value of 1+ cosec thita upon 1-sin thita ​

Answer 1

Question:-

If Cosθ = √3/2 then find the value of (1 + cosecθ)/(1 - secθ)

Given:-

• $\sf{Cos\theta = \dfrac{\sqrt{3}}{2}}$

To Find:-

• $\sf{The\:value\:of\:\dfrac{1 + cosec\theta}{1 - cosec\theta}}$

Solution:-

We have:-

• $\sf{Cosec\theta = \dfrac{\sqrt{3}}{2}}$

We know,

$\sf{Cos30^\circ = \dfrac{\sqrt{3}}{2}}$

Hence,

We can write:-

• Cosθ = Cos30°

=> θ = 30°

∴ The value of θ is 30°

Now,

We need to find the value of:-

$\sf{\dfrac{1 + cosec\theta}{1 - sec\theta}}$

Putting the value of θ

$\sf{ = \dfrac{1 + cosec30^\circ}{1 - sec30^\circ}}$

From trigonometric table we know,

• Cosec30° = 2
• $\sf{Sec30^\circ = \dfrac{2}{\sqrt{3}}}$

Hence putting the values:-

$\sf{ = \dfrac{1 + 2}{1 - \dfrac{2}{\sqrt{3}}}}$

$\sf{ = \dfrac{3}{\dfrac{\sqrt{3} - 2}{\sqrt{3}}}}$

$\sf{ = \dfrac{3\sqrt{3}}{\sqrt{3} -2}}$

By rationalising the denominator:-

$\sf{ = \dfrac{(3\sqrt{3})(\sqrt{3} + 2)}{(\sqrt{3} - 2)(\sqrt{3} + 2)}}$

$\sf{ = \dfrac{9 - 6\sqrt{3}}{(\sqrt{3})^2 - (2)^2}}$

$\sf{ = \dfrac{9 - 6\sqrt{3}}{3 - 4}}$

$\sf{ = \dfrac{9 - 6\sqrt{3}}{-1}}$

$\sf{ = 6\sqrt{3} - 9}$

∴ The value of $\sf{\dfrac{1 + cosec\theta}{1 - sec\theta}}$ is 6√3 - 9.

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Additional Information:-

Trigonometric table:-

$\boxed{\begin{array}{c|c|c|c|c|c} \theta & \sf{0^{\circ}} & \sf{30^{\circ}} & \sf{45^{\circ}} & \sf{60^{\circ}} & \sf{90^{\circ}} \\ \dfrac{\qquad \qquad}{} & \dfrac{\qquad \qquad}{} & \dfrac{\qquad \qquad}{} & \dfrac{\qquad \qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad \qquad}{} \\ \sf{Sin\theta} & \sf{0} & \sf{\dfrac{1}{2}} & \sf{\dfrac{1}{\sqrt{2}}} & \sf{\dfrac{\sqrt{3}}{2}} & \sf{1} \\ \sf{Cos\theta} & \sf{1} & \sf{\dfrac{\sqrt{3}}{2}} & \sf{\dfrac{1}{\sqrt{2}}} & \sf{\dfrac{1}{2}} & \sf{0} \\ \sf{Tan\theta} & \sf{0} & \sf{\dfrac{1}{\sqrt{3}}} & \sf{1} & \sf{\sqrt{3}} & \sf{Not\:defined} \\ \sf{Cosec\theta} & \sf{Not\:Defined} & \sf{2} & \sf{\sqrt{2}} & \sf{\dfrac{2}{\sqrt{3}}} & \sf{1} \\ \sf{Sec\theta} & \sf{1} & \sf{\dfrac{2}{\sqrt{3}}} & \sf{\sqrt{2}} & \sf{2} & \sf{Not\:Defined} \\ \sf{Cot\theta} & \sf{Not\:Defined} & \sf{\sqrt{3}} & \sf{1} & \sf{\dfrac{1}{\sqrt{3}}} & \sf{0} \end{array}}$

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Answer 2

$\huge \sf{ \underline{ \underline{Question:-}}}$

$\sf \: If Cos0= \frac{√ 3}{2} \: then \: find \: the \: value \: of \frac{ (1 + cosec0)}{(1 - sec0)}$

$\huge{ \bf{ \color{gold}{Given:-}}}$

• $\sf{Cos \theta = \dfrac{\sqrt{3}}{2}}$

$\huge{ \sf{ \color{purple}{To \: Find:-}}}$

$\sf{value\:of\:\dfrac{1 + cosec\theta}{1 - cosec\theta}}$

Solution:-

$\boxed{\bf{Cosec\theta = \dfrac{\sqrt{3}}{2}}}$

Cos30° = √3/2

Hence,

Cosθ = Cos30°

= θ = 30°

∴ The value of θ is 30°

We need to find the value of:-

$\bf{\dfrac{1 + cosec\theta}{1 - sec\theta}}$

• Putting the value of θ

$\bf{ = \dfrac{1 + cosec30^\circ}{1 - sec30^\circ}}$

• From trigonometric table we know,

=Cosec30° = 2

$\bf{Sec30^\circ = \dfrac{2}{\sqrt{3}}}$

• putting the values:-

$\bf{ = \dfrac{1 + 2}{1 - \dfrac{2}{\sqrt{3}}}}$

$\bf{ = \dfrac{3}{\dfrac{\sqrt{3} - 2}{\sqrt{3}}}}$

$\bf{ = \dfrac{3\sqrt{3}}{\sqrt{3} -2}}$

rationalising the denominator:-

$\bf{ = \dfrac{(3\sqrt{3})(\sqrt{3} + 2)}{(\sqrt{3} - 2)(\sqrt{3} + 2)}}$

$\bf{ = \dfrac{9 - 6\sqrt{3}}{(\sqrt{3})^2 - (2)^2}}$

$\bf{ = \dfrac{9 - 6\sqrt{3}}{3 - 4}}$

$\bf{ = \dfrac{9 - 6\sqrt{3}}{-1}}$

$\bf{ = 6\sqrt{3} - 9}$

$\sf \: ∴ The \: value \: of {\dfrac{1 + cosec\theta}{1 - sec\theta}}is \: 6√3 - 9.$

Learn More✔✔

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