Question

tan 10°tan 25° tan 45°tan 65° tan 80°=1​

Answer 1

ANSWER:

To Prove:

• tan 10° tan 25° tan 45° tan 65° tan 80° = 1

Proof:

$\text{We need to prove that,}\\\\:\longrightarrow\tan10^{\circ}\times\tan25^{\circ}\times\tan45^{\circ}\times\tan65^{\circ}\times\tan80^{\circ}=1\\\\\text{Solving LHS,}\\\\:\implies\tan10^{\circ}\times\tan25^{\circ}\times\tan45^{\circ}\times\tan65^{\circ}\times\tan80^{\circ}\\\\\text{We know that,}\\\\:\hookrightarrow\tan\theta=\cot(90-\theta)\\\\\text{So,}\\\\:\implies\tan10^{\circ}\times\tan25^{\circ}\times\tan45^{\circ}\times\cot(90-65)^{\circ}\times\tan(90-80)^{\circ}$

$:\implies\tan10^{\circ}\times\tan25^{\circ}\times\tan45^{\circ}\times\cot25^{\circ}\times\cot10^{\circ}\\\\\text{On rearranging,}\\\\:\implies(\tan10^{\circ}\times\cot10^{\circ})\times(\tan25^{\circ}\times\cot25^{\circ})\times\tan45^{\circ}\\\\\text{We know that,}\\\\:\hookrightarrow\tan\theta\times\cot\theta=1\\\\\text{So,}\\\\:\implies(\tan10^{\circ}\times\cot10^{\circ})\times(\tan25^{\circ}\times\cot25^{\circ})\times\tan45^{\circ}\\\\:\implies(1)\times(1)\times\tan45^{\circ}\\\\:\implies\tan45^{\circ}$

$\text{We know that,}\\\\:\hookrightarrow\tan45^{\circ}=1\\\\\text{So,}\\\\:\implies\tan45^{\circ}\\\\:\implies\bf{1=RHS}\\\\\text{\bf{HENCE PROVED!!!}}$

Formulae Used:

• tanA = cot(90 - A)
• tanA × cotA = 1
• tan45° = 1

Learn More:

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

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

Answer 2

Answer:

Good morning

sis

how are you