Question

evaluate tan(15°)-cot (15°)​

Answer 1

ᴀɴsᴡᴇʀ ʀᴇғᴇʀs ᴛᴏ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ.

Answer 2

ANSWER:

To Evaluate:

• tan(15)° - cot(15)°

Solution:

$\text{We need to find value of,}\\\\:\longrightarrow\tan15^{\circ}-\cot15^{\circ}\\\\\text{We know that,}\\\\:\hookrightarrow\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\:\hookrightarrow\cot\theta=\dfrac{\cos\theta}{\sin\theta}\\\\\text{So,}\\\\:\implies\dfrac{\sin15^{\circ}}{\cos15^{\circ}}-\dfrac{\cos15^{\circ}}{\sin15^{\circ}}\\\\:\implies\dfrac{\sin^215^{\circ}-\cos^215^{\circ}}{\sin15^{\circ}\,\cos15^{\circ}}$

$:\implies\dfrac{-(\cos^215^{\circ}-\sin^215^{\circ})}{\sin15^{\circ}\,\cos15^{\circ}}\\\\\text{We know that,}\\\\:\hookrightarrow\cos^2\theta-\sin^2\theta=\cos2\theta\\\\\text{So,}\\\\:\implies\dfrac{-[\cos2(15^{\circ})]}{\sin15^{\circ}\,\cos15^{\circ}}\\\\\text{Multiplying and dividing by 2,}\\\\:\implies\dfrac{-2[\cos30^{\circ}]}{2\sin15^{\circ}\,\cos15^{\circ}}$

$\text{We know that,}\\\\:\hookrightarrow2\sin\theta\cos\theta=\sin2\theta\\\\\text{So,}\\\\:\implies\dfrac{-2\cos30^{\circ}}{\sin2(15^{\circ})}\\\\:\implies\dfrac{-2\cos30^{\circ}}{\sin30^{\circ}}\\\\:\implies-2\times\dfrac{\cos30^{\circ}}{\sin30^{\circ}}\\\\:\implies-2\times\cot30^{\circ}\\\\\text{We know that,}\\\\:\hookrightarrow\cot30^{\circ}=\sqrt3\\\\\text{So,}\\\\:\implies-2\times\sqrt3\\\\\bf{:\implies-2\sqrt3}$

Formulae Used:

$:\hookrightarrow1)\:\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\:\hookrightarrow2)\:\cot\theta=\dfrac{\cos\theta}{\sin\theta}\\\\:\hookrightarrow3)\:\cos^2\theta-\sin^2\theta=\cos2\theta\\\\:\hookrightarrow4)\:2\sin\theta\cos\theta=\sin2\theta\\\\:\hookrightarrow5)\:\cot30^{\circ}=\sqrt3$

Learn More:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$