Question

5. Prove that 2tan80° = tan85° -tan 5°​

Answer 1

$\large \underline \bold{Solution}$:-

$\rm{\: \: \: \: \: \: \: \: \: tan \: 80}$

➱ $\rm{tan(85 \: - \: 5)}$

$\small \underline \bold{We \: know \: that}$:-

$\rm{tan(A - B) \: = \: \dfrac{tanA \: - \: tanB}{1 \: + \: tanA tanB}}$

$\rm{So \: ,}$

➱ $\rm{\dfrac{tan85 \: - \: tan5}{1 \: + \: tan85 tan5}}$

➱ $\rm{\dfrac{tan85 \: - \: tan5}{1 \: + \: tan(90 - 5) tan5}}$

As we know -

$\rm{tan(90 \: - \: \theta) \: = \: Cot \theta}$

$\rm{then \: ,}$

➱ $\rm{\dfrac{tan85 \: - \: tan5}{1 \: + \: Cot5 tan5}}$

now ,

$\rm{tan \theta\times Cot \theta \: = \: 1}$

$\rm{So \: - }$

➱ $\rm{\dfrac{tan 85 \: - \: tan5}{1 + 1}}$

➱ $\rm{\dfrac{tan85 \: - \: tan5}{2}}$

$\rm{Now \: ,}$

$\rm{as \: we \: got \: So \: , \: we \: can \: say \: -}$

➱ $\rm{tan80 \: = \: \dfrac{(tan85 \: - \: tan5)}{2}}$

➱ $\rm{2tan80 \: = \: (tan85 \: - \: tan5)}$

$\large \underline \bold{HENCE \: PROOF}$

Answer 2

Answer:

\large \underline \bold{Solution}

\rm{tan(A - B) \: = \: \dfrac{tanA \: - \: tanB}{1 \: + \: tanA tanB}}

Math

5 points

2.0

32

:-

\rm{\: \: \: \: \: \: \: \: \: tan \: 80}

\rm{tan(85 \: - \: 5)}

➱

\small \underline \bold{We \: know \: that}

:-

\rm{So \: ,}

\rm{\dfrac{tan85 \: - \: tan5}{1 \: + \: tan85 tan5}}

➱

\rm{\dfrac{tan85 \: - \: tan5}{1 \: + \: tan(90 - 5) tan5}}

➱

As we know -

\rm{tan(90 \: - \: \theta) \: = \: Cot \theta}

\rm{then \: ,}

\rm{\dfrac{tan85 \: - \: tan5}{1 \: + \: Cot5 tan5}}

➱

now ,

\rm{tan \theta\times Cot \theta \: = \: 1}

\rm{So \: - }

ANSWER

\rm{\dfrac{tan 85 \: - \: tan5}{1 + 1}}

➱