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T
=
b
2
2
(
cot
α
−
cot
(
α
+
β
)
)
=
b
2
sin
α
sin
(
α
+
β
)
2
sin
β
{\displaystyle T={\frac {b^{2}}{2(\cot \alpha -\cot(\alpha +\beta ))}}={\frac {b^{2}\sin \alpha \sin(\alpha +\beta )}{2\sin \beta }}}
Citation List of trigonometric identities#Sum-to-product identities
Proof
tan
θ
−
tan
φ
=
sin
(
θ
−
φ
)
cos
θ
cos
φ
{\displaystyle \tan \theta -\tan \varphi ={\frac {\sin(\theta -\varphi )}{\cos \theta \cos \varphi }}}
Changing
θ
→
π
2
−
α
,
φ
→
π
2
−
(
α
+
β
)
{\displaystyle \theta \rightarrow {\frac {\pi }{2}}-\alpha ,\varphi \rightarrow {\frac {\pi }{2}}-(\alpha +\beta )}
tan
(
π
2
−
α
)
−
tan
(
π
2
−
(
α
+
β
)
)
=
cot
α
−
cot
(
α
+
β
)
=
sin
(
−
α
−
(
−
(
α
+
β
)
)
)
cos
(
π
2
−
α
)
cos
(
π
2
−
(
α
+
β
)
)
=
sin
β
sin
α
sin
(
α
+
β
)
{\displaystyle \tan({\frac {\pi }{2}}-\alpha )-\tan({\frac {\pi }{2}}-(\alpha +\beta ))=\cot \alpha -\cot(\alpha +\beta )={\frac {\sin(-\alpha -(-(\alpha +\beta )))}{\cos({\frac {\pi }{2}}-\alpha )\cos({\frac {\pi }{2}}-(\alpha +\beta ))}}={\frac {\sin \beta }{\sin \alpha \sin(\alpha +\beta )}}}
So,
T
=
b
2
2
(
cot
α
−
cot
(
α
+
β
)
)
=
b
2
sin
α
sin
(
α
+
β
)
2
sin
β
{\displaystyle T={\frac {b^{2}}{2(\cot \alpha -\cot(\alpha +\beta ))}}={\frac {b^{2}\sin \alpha \sin(\alpha +\beta )}{2\sin \beta }}}
240D:1E:309:5F00:CAD2:3562:9725:7AEE (talk ) 02:11, 13 May 2025 (UTC) [ reply ]
240d:1e:309:5f00:21c1:5849:ba12:2272 , possibly the same person) to:[...] Citation Area of a triangle#Knowing ASA (angle-side-angle)
Proof
Since
cot
γ
=
−
cot
(
π
−
γ
)
=
−
cot
(
α
+
β
)
{\displaystyle \cot \gamma =-\cot(\pi -\gamma )=-\cot(\alpha +\beta )}
S
=
b
2
2
(
cot
α
+
cot
γ
)
=
b
2
2
(
cot
α
−
cot
(
α
+
β
)
)
{\displaystyle S={\frac {b^{2}}{2(\cot \alpha +\cot \gamma )}}={\frac {b^{2}}{2(\cot \alpha -\cot(\alpha +\beta ))}}}
In either case, I'm not sure what this comment is getting at. Maybe you can be more explicit about what you think should be changed in this article, and how? –jacobolus (t) 00:32, 4 June 2025 (UTC) [ reply ] 
