Question

1 Using tan(A + B) =
tan A+tan B
divided by
1 -tan A tan B
show that tan 75°=2+ V3.​

Answer 1

we know that

tan(A + B) = $\frac{tanA + tanB}{1 - tanAtanB}$

now,

$tan {75}^{o} \\ = tan( {30}^{o} + {45}^{o} ) \\ = \frac{tan {30}^{o} + tan {45}^{o} }{1 - tan {30}^{o} tan {45}^{o} } \\ = \frac{ \frac{1}{ \sqrt{3} } + 1}{1 - ( \frac{1}{ \sqrt{3} })(1) } \\ = \frac{ \frac{ \sqrt{3} + 1 }{ \sqrt{3} } }{ \frac{ \sqrt{3} - 1}{ \sqrt{3} } } \\ = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1} \\ = \frac{( \sqrt{3} + 1)( \sqrt{3} + 1)}{( \sqrt{3} - 1)( \sqrt{3} + 1)} \\ = \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} = \frac{4 + 2 \sqrt{3} }{2} \\ = 2 + \sqrt{3}$

...Hence Proved!

Answer 2

by using the given formula......we can solve this problem...here u go

tan(A+B)=tanA+tanB/1-tanAtanB

now....let us assume A=30° and B=45°...because here we asked to prove for 75° ok.....

next..... substitute the vaules in the above formula we get....

tan(30°+45°)=tan30°+tan45°/1-tan30°tan45°

now... substitute the vaules of tan30° and tan 45° vaules in above formula....we get

tan75°=(1/√3+1)/1-1/√3

on solving this we get....

1+√3/√3-1

now....rationalize the denominator....that is √3+1 and multiply this with both numerator and denominator....then we get

4+2√3/2 ...on solving this we get 2+√3....

this is ur required answer..... hence...prooved