tan (1/2)[ sin⁻¹( 2x/1+x² )+ cos⁻¹( 1-y² /1+y²) ], | x | < 1, y > 0 तथा xy < 1 का मान ज्ञात कीजिए
Answers
Given : tan (1/2)[ sin⁻¹( 2x/1+x² )+ cos⁻¹( 1-y² /1+y²) ], | x | < 1, y > 0 तथा xy < 1
To find : मान ज्ञात कीजिए
Solution:
tan (1/2)[ sin⁻¹( 2x/1+x² )+ cos⁻¹( 1-y² /1+y²) ]
sin⁻¹( 2x/1+x² )
माना x = tanα => α = tan⁻¹x
= sin⁻¹ (2tanα/(1 + tan²α))
हमें ज्ञात है की 1 + tan²α = sec²α
= sin⁻¹ (2tanα/sec²α)
1/sec²α = cos²α
= sin⁻¹ (2tanαcos²α)
= sin⁻¹ (2tanαcos αcosα)
tanαcos α = sinα
= sin⁻¹ (2sinα cosα)
= sin⁻¹ ( sin2α )
= 2α
= 2tan⁻¹x
sin⁻¹( 2x/1+x² ) = 2tan⁻¹x
cos⁻¹( (1-y²) /(1+y²) )
y = tanβ => β = tan⁻¹y
= cos⁻¹( (1-tan²β) /(1+tan²β) )
= cos⁻¹( (1-sin²β/cos²β) /(1+sin²β/cos²β) )
= cos⁻¹( (cos²β-sin²β) /(cos²β+sin²β) )
cos²β-sin²β = cos2β
cos²β+sin²β = 1
= cos⁻¹ (cos2β)
=2β
= 2tan⁻¹y
cos⁻¹( (1-y²) /(1+y²) ) = 2tan⁻¹y
tan (1/2)[ sin⁻¹( 2x/1+x² )+ cos⁻¹( 1-y² /1+y²) ]
sin⁻¹( 2x/1+x² ) = 2tan⁻¹x , cos⁻¹( (1-y²) /(1+y²) ) = 2tan⁻¹y
= tan(1/2) [ 2tan⁻¹x + 2tan⁻¹y ]
= tan [ tan⁻¹x + tan⁻¹y ]
tan ( A + B) = (TanA + TanB)/(1 - TanATanB)
= (Tan( tan⁻¹x) + Tan( tan⁻¹y) )/( 1 - Tan( tan⁻¹x) Tan( tan⁻¹y) )
= (x + y)/(1 - xy)
tan (1/2)[ sin⁻¹( 2x/1+x² )+ cos⁻¹( 1-y² /1+y²) ] का मान = (x + y)/(1 - xy)
और सीखें
tan⁻¹(2/11) + tan⁻¹ (7/24) = tan⁻¹ ½
https://brainly.in/question/16554897
3cos⁻¹ x = cos⁻¹ (4x³ - 3x), x∈[1/2, 1 ] को सिद्ध कीजिये
https://brainly.in/question/16554602
"cos⁻¹(-1/2 ) का मुख्य मान ज्ञात कीजिये"
brainly.in/question/16549994
cos⁻¹ (½) + 2sin⁻¹(½) का मान ज्ञात कीजिये
brainly.in/question/16554597
sin⁻¹(-1 / 2) का मुख्य मान ज्ञात कीजिये
brainly.in/question/16549991
Answer:
answer is above dude.
Thank you.