Explanation
tan−1(tan5π4)=α
⇒5π4=α
Also, tan−1(tan−2π3)=β
⇒−2π3=β
⇒α+β=5π4−2π3
⇒α+β=7π12
sin−1x+cos−1x
sin−1x=θ1⇒sinθ1=x
cos−1x=θ2⇒cosθ2=x
tanθ1=x√1−x2
tanθ2=√1−x2x
tan(θ1+θ2)=tanθ1+tanθ21−tanθ1tanθ2
=x√1−x2+√1−x2x1−1
=x2+1−x2x√1−x2(0)
tan(θ1+θ2)=∞
θ1+θ2=π2
∴sin−1x+cos−1x=π2
Let,
t=2tan−1[√a−ba+btanθ2] ……. (1)
b=acosα ……. (2)
Then,
a−ba+b=a−acosαa+acosα
a−ba+b=1−cosα1+cosα
a−ba+b=1−(1−2sin2α2)1+2cos2α2
a−ba+b=2sin2α22cos2α2
a−ba+b=tan2α2
√a−ba+b=tanα2 …… (3)
From equation (1) and (3), we get
tant2=tanα2tanθ2
⇒cost=1−tan2t21+tan2t2
cost=1−tan2α2tan2θ21+tan2α2tan2θ2
cost=1−sin2α2cos2α2sin2θ2cos2θ21+sin2α2cos2α2sin2θ2cos2θ2
cost=cos2α2cos2θ2−sin2α2sin2θ2cos2α2cos2θ2−sin2α2sin2θ2
cost=12[cosθ+cosα]12[1+cosαcosθ]
From equation (2), we get
cost=cosθ+ba1+cosθba=acosθ+ba+bcosx
⇒t=cos−1[acosθ+ba+bcosθ]
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