Math, asked by Garvnanwani3113, 9 months ago

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

Answered by amitnrw
0

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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