Question

Solve it with proper explanation guys.
i will be very thankful​

Answer 1

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

$\rm :\longmapsto\:tan15 \alpha - tan10\alpha - tan5\alpha = tan15\alpha \: tan10\alpha \: tan5\alpha$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

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

We know,

$\rm :\longmapsto\:15\alpha = 10\alpha + 5\alpha$

So,

$\rm :\longmapsto\:tan15\alpha =tan( 10\alpha + 5\alpha)$

Now, using identity,

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

we get,

$\rm :\longmapsto\:tan15\alpha = \dfrac{tan10\alpha + tan5\alpha}{1 - tan10\alpha \: tan5\alpha}$

$\rm :\longmapsto\:tan15\alpha - tan15\alpha \: tan10\alpha \: tan5\alpha = tan10\alpha + tan5\alpha$

Now, On transposition the terms, we get

$\rm :\longmapsto\:tan15 \alpha - tan10\alpha - tan5\alpha = tan15\alpha \: tan10\alpha \: tan5\alpha$

Hence, Proved.

Additional Information :-

Let's solve one more problem of same type!!

Question :- Prove that

$\rm :\longmapsto\:tan15 \degree \: + tan30\degree + tan15\degree tan30\degree = 1$

Solution :-

We know that

$\rm :\longmapsto\:45\degree = 30\degree + 15\degree$

So,

$\rm :\longmapsto\:tan45\degree =tan( 30\degree + 15\degree )$

$\rm :\longmapsto\:1 = \dfrac{tan15\degree + tan30\degree }{1 - tan15\degree tan30\degree }$

$\rm :\longmapsto\:tan15\degree + tan30\degree = 1 - tan15\degree tan30\degree$

On transposition the terms, we get

$\rm :\longmapsto\:tan15 \degree \: + tan30\degree + tan15\degree tan30\degree = 1$

Hence, Proved

Additional Formula :-

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

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

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

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

$\red{\boxed{ \bf{ \: tanx - tany = \frac{tanx - tany}{1 + tanxtany}}}}$