Solve this and plzz don't give any link Q 14 Let a, b, c be positive real numbers - Maths - Inverse Trigonometric Functions

Question

Solve this and plzz don't give any link
Q 14 Let a, b, c be positive real numbers Let θ = tan - 1 a (
a + b + c ) b c + tan - 1 b ( a + b + c ) c a + tan - 1 c ( a + b +
c ) a b Then tan θ =   _ _ _ _ _ _ _ _ _ _

Aarushi Mishra · Accepted Answer

tan x+y+z=tan x+tan y+tan z-tan x tan y tan z1-tan x tan y+tan x tan z+tan y tan zθ=tan-1aa+b+cbc+tan-1ba+b+cac+tan-1ca+b+cabθ=tan-1a2a+b+cabc+tan-1b2a+b+cabc+tan-1c2a+b+cabcθ=tan-1aa+b+cabc+tan-1ba+b+cabc+tan-1ca+b+cabctan θ=tan tan-1aa+b+cabc+tantan-1ba+b+cabc+tantan-1ca+b+cabc-tan tan-1aa+b+cabc tantan-1ba+b+cabc tantan-1ca+b+cabcsome terms in denominator=aa+b+cabc+ba+b+cabc+ca+b+cabc-aa+b+cabc ba+b+cabc ca+b+cabcsome terms in denominator=a+b+ca+b+cabc-abca+b+c3abc3some terms in denominator=a+b+c32abc-abca+b+c32abc32some terms in denominator=a+b+c32abc-a+b+c32abcsome terms in denominator=0

Solve this and plzz don't give any link. Q.14. Let a, b, c be positive real numbers. Let θ = tan - 1 a ( a + b + c ) b c + tan - 1 b ( a + b + c ) c a + tan - 1 c ( a + b + c ) a b . Then tan θ = _ _ _ _ _ _ _ _ _ _

Solve this and plzz don't give any link.
Q.14. Let a, b, c be positive real numbers. Let $θ = \tan^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + \tan^{- 1} \sqrt{\frac{b (a + b + c)}{c a}} + \tan^{- 1} \sqrt{\frac{c (a + b + c)}{a b}}$ . Then $\tan θ =__________$