Question

Can anyone solve this...

Answer 1

heya,
it is very simple yrr.

cosA=b/h=1/2

p=√h^2-b^2

p=√2^2-1^2,then tanA=p/b=√3/1

p=√3

again sinB=p/h

sinB=1/√2=p/h

b=√2^2-1^2
b=1

now tanB=p/b=√2/1

now putting on ,TanA+TanB/1-TanA+TanB=Tan(A+B).

hope you can find it

@rajukumar..

Answer 2

hey friend !

=)here given :-

cosA = 1/2 -------(1)

we know that cos60 = 1/2

so A = 60°

similarly , sinB = 1/√2

therefore , B = 45°

=) put these value in the required expression we get

$\frac{ \tan(a) - \tan(b) }{1 + \tan(a) \tan(b) ?}$

$\frac{ \tan(60) - \tan(45) }{1 + \tan(60) \tan(45) } = \frac{ \sqrt{3 } \: - 1}{1 + \sqrt{3} }$
so rationalise the above expression we get ...

◆(√3 -1)^2 /3-1 = (3 + 1 -2√3)/2

=) ( 2 -√3 ) answer