Question

if A,B,C are the angles of a triangle, show that cot B = cot (A + 2B + C)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• A,B,C are the angles of a triangle.

We know that,

• Sum of interior angles of a triangle is 180°.

$\bf\implies \:A + B + C = 180\degree$

Now,

Consider,

$\rm :\longmapsto\:cot(A + 2B + C)$

$\rm \: = \: \: \:cot(A + B + B + C)$

$\rm \: = \: \: \:cot(A + B + C + B)$

$\rm \: = \: \: cot(180\degree + B)$

$\rm \: = \: \: cot(90\degree \times 2 + B)$

$\rm \: = \: \: cotB$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ