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show that 2tan^-1[tan(45-a)tanb/2] = cos^-1[(sin2a+cosb)/(!+sin2acosb)]

show that 2tan^-1[tan(45-a)tanb/2] = cos^-1[(sin2a+cosb)/(!+sin2acosb)]

Grade:12

1 Answers

Vikas TU
14149 Points
one year ago
Let z = 2tan⁻¹( tan(45−a)tan(b/2) ) so tan²(z/2) = tan²(45−a)tan²(b/2)
tan²(45−a) = (1−tana)²/(1+tana)² = (cosa−sina)²/(cosa+sina)² = (1−sin2a)/(1+sin2a)
tan²(b/2) = sin²(b/2)/cos²(b/2) = (1−cosb)/(1+cosb)
cosz = (1−tan²(z/2))/(1+tan²(z/2)) = 
{ (1+cosb)(1+sin2a)–(1−sin2a)(1−cosb) } / { (1+cosb)(1+sin2a)+(1−sin2a)(1−cosb) }
      Num = 1+sin2a+cosb+cosbsin2a − 1+cosb+sin2a−sin2acosb = 2(sin2a+cosb)
      Den = 1+sin2a+cosb+cosbsin2a + 1−cosb−sin2a+sin2acosb = 2(1+sin2acosb)
      ∴ cosz = (sin2a+cosb)/(1+sin2acosb)

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