Question

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

T.P.T,
$1 + \cot \alpha \tan \beta = 0$

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Answer 1

i hope this will help you

Answer 2

Solution:

Given $\sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0$ ----(1)

Multiply both sides by (-1), we get

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

$\implies cos(\alpha+\beta)-1=0$

/* Since ,

Cos(A+B) = cosAcosB-sinAsinB */

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

$\implies cos(\alpha+\beta)=cos0$

$\implies \alpha+\beta = 0$---(1)

Now ,

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

= $1+\frac{cos\alpha}{sin\alpha}\times\frac{sin\beta}{cos\beta}$

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

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

/* Since,

sinAcosB+cosAsinB = sin(A+B)*/

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

/* from (1) */

= 0

= RHS

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