sin (tan⁻¹ ), |x|<1 बराबर होता हे :
(A) x/ √1-x² (B) 1/ √1-x² (C) 1/ √1+x² (D) x/ √1+x²
Answers
Given : sin (tan⁻¹ x) |x|<1
To find : सही विकल्प चुनें
Solution:
माना sin (tan⁻¹ x) = y
=> tan⁻¹ x = sin⁻¹y
माना tan⁻¹ x = A
=> tanA = x
Sec²A = 1 + tan²A
=> sec²A = 1 + x²
=> secA = √(1 + x²)
=> CosA = 1/√(1 + x²)
SinA = CosA * TanA
=> SinA = x/√(1 + x²)
=> A = Sin⁻¹(x/√(1 + x²))
=> tan⁻¹ x = Sin⁻¹(x/√(1 + x²))
=> Sin ( tan⁻¹ x) = Sin (Sin⁻¹(x/√(1 + x²)))
=> Sin ( tan⁻¹ x) = x/√(1 + x²)
सही विकल्प (D) x/ √1+x²
दोनों तरफ Tan लेने पर
=> Tan ( 2tan⁻¹((1-x)/(1+x)) = Tan (tan⁻¹ ( x))
Tan2α = 2Tanα/(1 - Tan²α)
α = tan⁻¹((1-x)/(1+x)
=> 2(1-x)/(1+x) /( 1 - ((1-x)/(1+x))²) = x
= 2(1 - x)(1 + x) / ( (1 + x)² -(1 - x)² ) = x
=> 2 (1 - x²)/( 4x) = x
=> 1 - x² = 2x²
=> 3x² = 1
=> x² = 1/3
=> x = 1/√3 ∵ x > 0
x = 1/√3
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