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If \frac{1}{\cos 290^{\circ}} +\frac{1}{\sqrt{3} \sin 250^{\circ}}=\lambda then the value of 9 \lambda^4+81 \lambda^2+97 must be 

Option: 1

715


Option: 2

726


Option: 3

785 


Option: 4

720


Answers (1)

best_answer

Here, \cos 290^{\circ}=\cos \left(270^{\circ}+20^{\circ}\right)=\sin 20^{\circ}

and    \sin 250^{\circ}=\sin \left(270^{\circ}-20^{\circ}\right)=-\cos 20^{\circ}

\therefore  The given expression  \frac{1}{\sin 20^{\circ}}-\frac{1}{\sqrt{3} \cos 20^{\circ}}=\lambda

\begin{aligned} & \Rightarrow \quad \frac{1}{\sin 20^{\circ}}-\frac{\cos 60^{\circ}}{\sin 60^{\circ} \cos 20^{\circ}}=\lambda \\ \\& \Rightarrow \quad \frac{\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ} \sin 60^{\circ}}=\lambda \end{aligned}

\Rightarrow \quad \frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\frac{\sin 40^{\circ}}{2} \times \frac{\sqrt{3}}{2}}=\lambda

\begin{aligned} & \therefore \quad \lambda=\frac{4}{\sqrt{3}} \\ \\& \Rightarrow \lambda^2=\frac{16}{3} \\ \end{aligned}

Then, 9 \lambda^4+81 \lambda^2+97=9 \times \frac{256}{9}+81 \times \frac{16}{3}+97

                                               \begin{aligned} & =256+432+97 \\ \\& =785 \end{aligned}

Posted by

Sumit Saini

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