Question

If $\sin\alpha \sin\beta-\cos\alpha \cos\beta+1=0$ , then prove that $1+\tan\beta \cot\alpha=0$​

Answer 1

Given,

LHS:-

$\sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0$

$\sin \alpha \sin \beta - \cos \alpha \cos \beta = - 1$

$- ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) = - 1$

$\cos \alpha \cos \beta - \sin \alpha \sin \beta = 1$

$\cos( \alpha + \beta ) = 1$

Now

To prove:-

RHS:-

$1 + \tan \alpha \cot \beta = 0$

$1 + \frac{ \tan\beta }{ \tan\alpha } = 0$

$\frac{ \tan \alpha + \tan\beta }{ \tan \alpha } = 0$

$\tan \alpha + \tan \beta = 0$

$\frac{ \sin \alpha }{ \ \cos \ \alpha } + \frac{ \sin \beta }{ \cos\beta } = 0$

$\frac{ \sin\alpha \cos \beta + \cos \alpha \sin \beta ) }{ \cos\alpha \cos\beta } = 0$

sin(a+ß) = 0.....(i)

Now,

let (a+ß) = x

we know,

sin²x+cos²x = 1

0+cos²x = 1 {from ep(i)}

cosx = 1

that is cos(a+ß) = 1 which I proved in LHS section.

Hence proved, LHS = RHS.

Cheers!

Answer 2

hence proved hope it helps