Question

tan x° = tan30°/ 1 + tan^2 30° [tan25°cot25°+ tan^2 45° / cot^2 60°]. Find x​

Answer 1

Required value of x is 60°

$\underline{\mathbb{EXPLANATION}}\bold{:}$

$\textsf{Given,}$

$\small{\mathsf{tanx^{\circ}=\frac{tan30^{\circ}}{1+tan^{2}30^{\circ}}\times \bigg[tan25^{\circ}\:cot25^{\circ}+\frac{tan^{2}45^{\circ}}{cot^{2}60^{\circ}}\bigg]}}$

$\small{\mathsf{=\frac{tan30^{\circ}}{1+tan^{2}30^{\circ}}\times \bigg[1+\frac{tan^{2}45^{\circ}}{cot^{2}60^{\circ}}\bigg]\:[\because tan\theta\:cot\theta=1]}}$

$\mathsf{=\frac{tan30^{\circ}}{1+tan^{2}30^{\circ}}\times\big[1+tan^{2}45^{\circ}\:tan^{2}60^{\circ}\big]}$

$\mathsf{=\frac{tan30^{\circ}}{1+tan^{2}30^{\circ}}\times\big[1+(tan45^{\circ}\:tan60^{\circ})^{2}\big]}$

$\mathsf{=\frac{\frac{1}{\sqrt{3}}}{1+(\frac{1}{\sqrt{3}})^{2}}\times\big[1+(1\times \sqrt{3})^{2}\big]}$

$\mathsf{=\frac{\frac{1}{\sqrt{3}}}{1+\frac{1}{3}}\times\big[1+(\sqrt{3})^{2}\big]}$

$\mathsf{=\frac{\frac{1}{\sqrt{3}}}{\frac{3+1}{3}}\times\big[1+3\big]}$

$\mathsf{=\frac{1}{\sqrt{3}}\times\frac{3}{4}\times 4}$

$\mathsf{=\sqrt{3}=tan60^{\circ}}$

$\implies \mathsf{tanx^{\circ}=tan60^{\circ}}$

$\implies \boxed{\bold{x=60^{\circ}}}$

$\therefore \mathsf{the </p><p>\:required\:value\:of\:x\:is\:60^{\circ}.}$

$\underline{\mathbb{TRIGONOMETRY}}\bold{:}$

$\quad\quad\quad \mathsf{1.\:tan30^{\circ}=\frac{1}{\sqrt{3}}}$

$\quad\quad\quad \mathsf{2.\:tan45^{\circ}=1}$

$\quad\quad\quad \mathsf{3.\:tan60^{\circ}=\sqrt{3}}$

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