Question

If A+B - π/2 then tan Atan B = 1 prove it​

Answer 1

Given:-

• A+B = $\frac{\pi}{2}$

To proove:-

• tanA tanB = 1

Proof:-

It is given that,

A+ B = $\frac{\pi}{2}$

:Now take (tan) on both side,

So,

$\tan(a + b) = \tan( \frac{\pi}{2} ) \\ \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } = \frac{1}{0}$

Now reverse the whole eqution,

$\frac{1 - \tan(a) \tan(b) }{ \tan(a) + \tan(b) } = \frac{0}{1}$

Now,

1- tanA tanB = 0

tanA tanB = 1

Hence proved

Hope its help uh❤

Answer 2

$\huge\underline\bold{AnSwEr,}$

Given:-

A+B = $\frac{\pi}{2}$

To proove:-

tanA tanB = 1

Proof:-

It is given that,

A+ B = $\frac{\pi}{2}$

:Now take (tan) on both side,

So,

$\tan(a + b) = \tan( \frac{\pi}{2} ) \\ \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } = \frac{1}{0}$

Now reverse the whole eqution,

$\frac{1 - \tan(a) \tan(b) }{ \tan(a) + \tan(b) } = \frac{0}{1}$

Now,

1- tanA tanB = 0

tanA tanB = 1

Hence proved

MasterHunTer\:\;/;/