Q If ϕ= tan-1x32K-xand θ = tan-12x-KK3then Prove that value of ϕ-θ= 30 - Maths - Inverse Trigonometric Functions

Question

Q   I f   ϕ =   tan - 1 x 3 2 K - x a n d
  θ   =     tan - 1 2 x - K K 3 t h e n
  P r o v e   t h a t   v a l u e   o f  
ϕ - θ =   30 °

Aarushi Mishra · Accepted Answer

ϕ=tan-1x32K-x and θ=tan-12x-KK3tan ϕ=x32K-x and tan θ=2x-KK3Considertan ϕ-θ=tan ϕ-tan θ1+tan ϕ tan θ=x32K-x-2x-KK31+x32K-x×2x-KK3=3Kx-4Kx+2K2+2x2-Kx2K-xK31+23x2-3Kx2K-xK3=3Kx-4Kx+2K2+2x2-Kx2K-xK32K-xK3+23x2-3Kx2K-xK3=3Kx-4Kx+2K2+2x2-Kx2K-xK323K2-3Kx+23x2-3Kx2K-xK3=2K2-2Kx+2x22K-xK323K2-23Kx+23x22K-xK3=2K2-2Kx+2x223K2-23Kx+23x2=132K2-2Kx+2x22K2-2Kx+2x2tan ϕ-θ=13= tan 30°ϕ-θ=30°

Q . I f ϕ = tan - 1 x 3 2 K - x a n d θ = tan - 1 2 x - K K 3 t h e n P r o v e t h a t v a l u e o f ϕ - θ = 30 °

$Q . I f ϕ = \tan^{- 1} \frac{x \sqrt{3}}{2 K - x} a n d θ = \tan^{- 1} \frac{2 x - K}{K \sqrt{3}} t h e n P r o v e t h a t v a l u e o f ϕ - θ = 30 °$