Question

If tan 69 + tan 66 − tan 66 tan 69 = 2, ℎ = a) -1 b) ½ c) – ½ d) None​

Answer 1

$\large\underline{\sf{To\:Find - }}$

The value of

$\rm :\longmapsto\:tan69\degree + tan66\degree - tan69\degree tan66\degree$

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

Consider,

$\rm :\longmapsto\:tan135\degree = tan(69\degree + 66\degree )$

We know,

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

So, using this

$\rm :\longmapsto\:tan(180\degree - 45\degree ) = \dfrac{tan69\degree + tan66\degree }{1 - tan69\degree tan66\degree }$

We know,

$\boxed{ \bf{ \: tan(180\degree - x) = - tanx}}$

So, using this, we get

$\rm :\longmapsto\: - \: tan( 45\degree ) = \dfrac{tan69\degree + tan66\degree }{1 - tan69\degree tan66\degree }$

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

$\rm :\longmapsto\: - 1 + tan69\degree tan66\degree = tan69\degree + tan66\degree$

$\bf\implies \:tan69\degree + tan66\degree - tan69\degree tan66\degree = - 1$

Hence,

• Option a) is correct.

Additional Information :-

$\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \bf{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \bf{ \: cos(x - y) = cosxcosy + sinxsiny}}$

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

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

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