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Sin⁻1(3/5) + tan⁻1(1/7) is equal to
- a)π/4
- b)π/2
- c)π
- d)cos⁻1(4/5)
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Sin1(3/5) + tan1(1/7) is equal toa)π/4b)π/2c)πd)cos1(4/5)Corr...
Solution:
Given:
Sin-1(3/5) + tan-1(1/7)
Using the identity:
tan-1(x) = sin-1(x / √(1 + x2))
Expressing tan-1(1/7) in terms of sin-1:
tan-1(1/7) = sin-1(1/7 / √(1 + (1/7)2))
tan-1(1/7) = sin-1(1/7 / √(1 + 1/49))
tan-1(1/7) = sin-1(1/7 / √(50/49))
tan-1(1/7) = sin-1(1/7 * 7/√50)
tan-1(1/7) = sin-1(1/√50)
Substitute back into the original expression:
Sin-1(3/5) + sin-1(1/√50)
Using the identity:
sin-1(x) + sin-1(y) = sin-1(x√(1-y2) + y√(1-x2))
Applying the identity:
sin-1(3/5√(1 - (1/√50)2) + (1/√50)√(1 - (3/5)2))
sin-1(3/5√(1 - 1/50) + (1/√50)√(1 - 9/25))
sin-1(3/5√(49/50) + (1/√50)√(16/25))
sin-1(3√49/5√50 + 4/5√50)
sin-1(3/5 + 4/5√50)
sin-1((3 + 4√50) / 5)
The angle whose sine is (3 + 4√50) / 5 is π/4.
Therefore, Sin-1(3/5) + tan-1(1/7) = π/4, which corresponds to option (a).
Given:
Sin-1(3/5) + tan-1(1/7)
Using the identity:
tan-1(x) = sin-1(x / √(1 + x2))
Expressing tan-1(1/7) in terms of sin-1:
tan-1(1/7) = sin-1(1/7 / √(1 + (1/7)2))
tan-1(1/7) = sin-1(1/7 / √(1 + 1/49))
tan-1(1/7) = sin-1(1/7 / √(50/49))
tan-1(1/7) = sin-1(1/7 * 7/√50)
tan-1(1/7) = sin-1(1/√50)
Substitute back into the original expression:
Sin-1(3/5) + sin-1(1/√50)
Using the identity:
sin-1(x) + sin-1(y) = sin-1(x√(1-y2) + y√(1-x2))
Applying the identity:
sin-1(3/5√(1 - (1/√50)2) + (1/√50)√(1 - (3/5)2))
sin-1(3/5√(1 - 1/50) + (1/√50)√(1 - 9/25))
sin-1(3/5√(49/50) + (1/√50)√(16/25))
sin-1(3√49/5√50 + 4/5√50)
sin-1(3/5 + 4/5√50)
sin-1((3 + 4√50) / 5)
The angle whose sine is (3 + 4√50) / 5 is π/4.
Therefore, Sin-1(3/5) + tan-1(1/7) = π/4, which corresponds to option (a).
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Sin1(3/5) + tan1(1/7) is equal toa)π/4b)π/2c)πd)cos1(4/5)Correct answer is option 'A'. Can you explain this answer?
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