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$\angle A$ અને $\angle B$ એવા લઘુકોણો છે કે, જેથી $\cos A =\cos B .$ સાબિત કરો કે $\angle A =\angle B$.
Solution

Let us consider a triangle $ABC$ in which $CD \perp AB$.
It is given that,
$\cos A=\cos B$
$\Rightarrow \frac{A D}{A C}=\frac{B D}{B C}$
$\Rightarrow \frac{A D}{B D}=\frac{A C}{B C}$
Let $\frac{A D}{B D}=\frac{A C}{B C}=k$
$\Rightarrow AD =k BD \ldots(1)$
And, $A C=k B C \ldots(2)$
Using Pythagoras theorem for triangles $CAD$ and $CBD,$ we obtain
$CD ^{2}= AC ^{2}- AD ^{2} \ldots(3)$
And, $CD ^{2}= BC ^{2}- BD ^{2} \ldots(4)$
From equations $( 3 )$ and $(4),$ we obtain
$AC ^{2}- AD ^{2}= BC ^{2}- BD ^{2}$
$\Rightarrow(k BC )^{2}-(k BD )^{2}= BC ^{2}- BD ^{2}$
$\Rightarrow k^{2}\left(B C^{2}-B D^{2}\right)=B C^{2}-B D^{2}$
$\Rightarrow k^{2}=1$
$\Rightarrow k=1$
Putting this value in equation $(2),$ we obtain
$AC = BC$
$\Rightarrow \angle A=\angle B$ (Angles opposite to equal sides of a triangle)