Math, asked by symashah000, 4 hours ago

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

Attachments:

Answers

Answered by mathdude500
2

\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}}}}

Similar questions