Question

if tan alpha =5/6 and tan beta =1/11 then prove that alpha+beta =Π/4

Answer 1

tan(a+b) = (tana+tanb)/(1-tana tanb)
tan(a+b) = (5/6+1/11)/(1-5/6×1/11)
= (55+6/66)/(1-5/66)
= (61/66)/(61/66)
= 1
tan(a+b) = tanπ/4
so,a+b= π/4

Answer 2

Answer:

$\tan ^{-1}\left(\frac{5}{6}\right)+\tan ^{-1}\left(\frac{1}{11}\right)=\alpha+\beta=\frac{\pi}{4}$  is proved

Step-by-step explanation:

We can write  

$\begin{array}{l}{\tan \alpha=\frac{5}{6} \quad \text { or } \quad \alpha=\tan ^{-1}\left(\frac{5}{6}\right)} \\ {\text { and } \tan \beta=\frac{1}{11} \text { or } \beta=\tan ^{-1}\left(\frac{1}{11}\right)}\end{array}$

If $\bold{\alpha+\beta=\frac{\pi}{4}}$ then in place of α and β we can put  

$\tan ^{-1}\left(\frac{5}{6}\right)+\tan ^{-1}\left(\frac{1}{11}\right)=\frac{\pi}{4}$

According to the inverse trigonometric formulae  

$\begin{array}{l}{\tan ^{-1}(x)+\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)} \\ {x=\frac{5}{6} \text { and } y=\frac{1}{11}} \\ {\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\tan ^{-1}\left(\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6} \cdot \frac{1}{11}}\right)=\tan ^{-1}(1)}\end{array}$

As we, all know $\tan ^{-1}(1)$  can be written as 45° or  π/4    

Hence, it is proved that $\bold{\tan ^{-1}\left(\frac{5}{6}\right)+\tan ^{-1}\left(\frac{1}{11}\right)=\alpha+\beta=\frac{\pi}{4}}$