Tan32+tan13+tan32tan13=1

tan 32^o + tan 13^o + tan 32^o.tan13^o = 1

tan 32^{<font size="2">o</font>} + tan 13^{<font size="2">o</font>} = 1 - tan 32^{<font size="2">o</font>}.tan13^{<font size="2">o</font>}

(tan 32^{<font size="2">o</font>} + tan 13<font size="2">^o)/(</font>1 - tan 32^{<font size="2">o</font>}.tan13^{<font size="2">o</font>}<font size="2">) = 1</font>

<font size="2">tan (32^o + 13^o) = 1                                        [Since tan (A + B) = (tan A + tan B)/(1 - tan A. tan B)]</font>

<font size="2">tan 45^o = 1</font>

<font size="2">1 = 1                                                            [Since tan 45^o = 1]</font>

<font size="2">Hence proved.</font>

Hello friends,

If we can recall the formula

tan( A+B)= ( tan A + tan B)  /(1- tan A tan B)

so let us take

A= 32 and B= 13

therefore   tan( 32+13) = (tan 32+ tan 13)/(1-tan32 tan13)

i.e    tan 45 = (tan 32+ tan 13)/(1-tan32 tan13)        [since 32+13 =45]

 1= (tan 32+ tan 13)/(1-tan32 tan13)       [tan 45 =1]

(1-tan32 tan13) = (tan 32+ tan 13)

1=  tan32 tan13 + tan 32+ tan 13

PROOVED

Thanks

S Mukherjee

7864927899