If tan 15^{circ}+frac{1}{tan 75^{circ}}+frac{ | vedclass

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Question

(A) Given the expression: $	an 15^{\circ} + \cot 75^{\circ} + \cot 105^{\circ} + 	an 195^{\circ} = 2a$.We know that $	an 15^{\circ} = 2-\sqrt{3}$.$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given the expression: $	an 15^{\circ} + \cot 75^{\circ} + \cot 105^{\circ} + 	an 195^{\circ} = 2a$.We know that $	an 15^{\circ} = 2-\sqrt{3}$.$\cot 75^{\circ} = 	an(90^{\circ}-75^{\circ}) = 	an 15^{\circ} = 2-\sqrt{3}$.$\cot 105^{\circ} = \cot(180^{\circ}-75^{\circ}) = -\cot 75^{\circ} = -(2-\sqrt{3}) = \sqrt{3}-2$.$	an 195^{\circ} = 	an(180^{\circ}+15^{\circ}) = 	an 15^{\circ} = 2-\sqrt{3}$.Substituting these values into the expression:$(2-\sqrt{3}) + (2-\sqrt{3}) + (\sqrt{3}-2) + (2-\sqrt{3}) = 2a$.Simplifying the left side:$2-\sqrt{3} + 2-\sqrt{3} + \sqrt{3}-2 + 2-\sqrt{3} = 4 - 2\sqrt{3} = 2a$.Thus,$a = 2-\sqrt{3}$.Now,calculate $a + \frac{1}{a}$:$a + \frac{1}{a} = (2-\sqrt{3}) + \frac{1}{2-\sqrt{3}}$.Rationalizing $\frac{1}{2-\sqrt{3}}$:$\frac{1}{2-\sqrt{3}} 	imes \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$.Therefore,$a + \frac{1}{a} = (2-\sqrt{3}) + (2+\sqrt{3}) = 4$.

If $\tan 15^{\circ}+\frac{1}{\tan 75^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2a$,then the value of $\left(a+\frac{1}{a}\right)$ is:

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