Question

If tan a=1 and tan b =√3 then cos a.cos b-sina.sinb

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If tan a=1 and tan b =√3 ,then ( cos a . cos b - sin a . sin b ) .

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\blue{\bf{Given}}}}}$

• tan a=1
• tan b =√3 .

$\Large{\underline{\mathfrak{\blue{\bf{Find}}}}}$

• ( cos a . cos b - sin a . sin b )

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

we know,

$\Large{\underline{\mathfrak{\blue{\bf{In\:\triangle\:ABC}}}}}$

$\small\boxed{\sf{\pink{\:\tan a\:=\:\dfrac{Perpendiculer}{Base}\:=\:\dfrac{AB}{BC}\:=\:\dfrac{1}{1}}}}$

$\Large{\underline{\mathfrak{\blue{\bf{By\:Pythagoras\:Theorem}}}}}$

$\small\boxed{\sf{\pink{\:(Hypotenuse)^2\:=\:(perpendiculer)^2+(Base)^2}}} \\ \\ \mapsto\sf{\:(Hypotenuse)^2\:=\:(1)^2+(1)^2} \\ \\ \mapsto\sf{\:(Hypotenuse)\:=\:\sqrt{2}}$

Now,

$\small\boxed{\sf{\pink{\:\cos a\:=\:\dfrac{Base}{Hypotenuse}\:=\:\dfrac{1}{\sqrt{2}}}}}$

And,

$\small\boxed{\sf{\pink{\:\sin a\:=\:\:\dfrac{perpendicular}{Hypotenuse}\:=\:\dfrac{1}{\sqrt{2}}}}}$

Again,

$\large{\underline{\mathfrak{\blue{\bf{In\:\triangle\:PQR}}}}}$

$\small\boxed{\sf{\pink{\:\tan b\:=\:\dfrac{perpendicular}{Base}\:=\:\dfrac{\sqrt{3}}{1}}}}$

$\large{\underline{\mathfrak{\blue{\bf{By\:Pythagoras\:Theorem}}}}}$

$\small\boxed{\sf{\pink{\:(Hypotenuse)^2\:=\:(perpendicdzqquler)^2+(Base)^2}}} \\ \\ \mapsto\sf{\:(Hypotenuse)^2\:=\:(\sqrt{3})^2+(1)^2} \\ \\ \mapsto\sf{\:(Hypotenuse)\:=\:\sqrt{4}} \\ \\ \mapsto\sf{\:(Hypotenuse)\:=\:2}$

Now,

$\small\boxed{\sf{\pink{\:\cos b\:=\:\dfrac{Base}{Hypotenuse}\:=\:\dfrac{1}{2}}}}$

and,

$\small\boxed{\sf{\pink{\:\sin b=\:\dfrac{perpendicular}{Hypotenuse}\:=\:\dfrac{\sqrt{3}}{2}}}}$

Now,Calculate

$\mapsto\sf{\orange{\:(\cos a . \cos b - \sin a . \sin b)}}$

( Keep value of cos a , cos b , sin a and sin b )

$\mapsto\sf{\:\dfrac{1}{\sqrt{2}}.\dfrac{1}{2}\:-\:\dfrac{1}{\sqrt{2}}.\dfrac{\sqrt{3}}{2}} \\ \\ \mapsto\sf{\:\dfrac{1}{4}\:-\:\dfrac{\sqrt{3}}{2\sqrt{2}}} \\ \\ \mapsto\sf{\:\dfrac{1}{4}\:-\:\dfrac{\sqrt{6}}{4}} \\ \\ \mapsto\sf{\blue{\:\dfrac{(1-\sqrt{6})}{4}\:\:\:\:\:\:\:Ans}}$

Answer 2

$\huge \tt \it \bf \it \bm { \mathbb{ \fcolorbox{blue}{yellow}{ \red{ANSWER :}}}} \green\longrightarrow</p><p>$

$\large { \boxed {\fbox{\frac{1 - \sqrt{6} }{4}}}}$

solution was in the attachment.