Question

anyone interested in trigonometrical solve it please


$if \: \sin\alpha - \sin \beta = \frac{1}{3}$
$\cos \beta - \cos \alpha = \frac{1}{2}$
$show \: that \:$
$\cot( \frac{ \alpha + \beta }{2} ) = \frac{2}{3}$
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Answer 1

Answer:

Step-by-step explanation:

Hey mate;

As you asked for I put it solution for you question;

For this I used following identities;

Sin C - Sin D = 2 cos { (C+D)÷2} × sin (C-D) ÷ 2;

Sin C - Sin D = 2 cos { (C+D)÷2} × sin (C-D) ÷ 2;Cos C - Cos D = 2 sin {(C+D) ÷ 2} × sin (D-C ) ÷ 2;

Sin C - Sin D = 2 cos { (C+D)÷2} × sin (C-D) ÷ 2;Cos C - Cos D = 2 sin {(C+D) ÷ 2} × sin (D-C ) ÷ 2;Or

Sin C - Sin D = 2 cos { (C+D)÷2} × sin (C-D) ÷ 2;Cos C - Cos D = 2 sin {(C+D) ÷ 2} × sin (D-C ) ÷ 2;Or -2SinC+D/2.sinC-D/2

.

After that;

In last second step I took

alpha+beta/2

alpha+beta/2as comman;

alpha+beta/2as comman;in cosalpha+beta/2

alpha+beta/2as comman;in cosalpha+beta/2sinalpha+beta/2

;

Other identity used=cosx/sinx

=cot x;

Hope you understand Hit like;

Be brainly

Answer 2

Answer:

Step-by-step explanation:

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