Question

tan 69° + tan 66° by 1 - tan 69º tan 66​

Answer 1

Given Question

Evaluate the following

$\rm :\longmapsto\:\dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree }$

$\green{\large\underline{\sf{Solution-}}}$

Given Trigonometric function is

$\rm :\longmapsto\:\dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany}}}}$

So, here

$\purple{\rm :\longmapsto\:x = 69\degree }$

and

$\purple{\rm :\longmapsto\:y = 66\degree }$

So, above expression can be rewritten as

$\rm \:  =  \: tan(69\degree + 66\degree )$

$\rm \:  =  \: tan135\degree$

can be further rewritten as

$\rm \:  =  \: tan(180\degree - 45\degree )$

$\rm \:  =  \: - tan45\degree$

$\rm \:  =  \: - \: 1$

Hence,

$\rm\implies \:\:\boxed{\tt{ \dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree } = - \: 1 \: }}$

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MORE TO KNOW

$\boxed{\tt{ sin(x + y) = sinx \: cosy + cosx \: siny \: }}$

$\boxed{\tt{ sin(x - y) = sinx \: cosy - cosx \: siny \: }}$

$\boxed{\tt{ cos(x + y) = cosx \: cosy \: - \: siny \: sinx \: }}$

$\boxed{\tt{ cos(x - y) = cosx \: cosy \: + \: siny \: sinx \: }}$

$\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany}}}$

$\boxed{\tt{ tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany}}}$

$\boxed{\tt{ tan(45\degree + x) = \frac{1 + tanx}{1 - tanx}}}$

$\boxed{\tt{ tan(45\degree - x) = \frac{1 - tanx}{1 + tanx}}}$

Answer 2

Question:-

Evaluate the Following:

$\\ \sf \dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree }$

Given:-

$\sf \longmapsto\:\dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree }$

To Solve:-

$\sf\longmapsto\:\dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree }$

Solution:-

Let, us consider the L.H.S.

$\sf \dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree }$

As we Known that,

$\sf \large tan (A+B)= \frac{(tanA + tanB)}{(1 - tanA.tanB)}$

Here,

• A = 69°.

• B = 66°

$\sf \therefore \dfrac{(tan69\degree + tan66\degree) }{(1 - tan69\degree \: tan66\degree) } = tan(69 + 66) \degree$

tan = 135°

– tan = 45°

= – 1.

= R.H.S.

$\sf \large \red{∴ L.HS. = R.HS.}$

$\sf \large \blue{Thus,Proved.}$

Answer:-

$\sf { \boxed{ \sf \green{\dfrac{tan69\degree + tan66\degree }{1 - tan69\degree \: tan66\degree } = - 1}}}$

Hope you have satisfied. ⚘