हिंदी

Prove that: tan-1 (sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x ) + sqrt(1 - x)) = π/4 - 1/2 cos-1 x, for -1/sqrt2 ≤ x ≤ 1 [Hint: Put x = cos 2θ] - Mathematics and Statistics

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प्रश्न

Prove that:

`tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x))) = pi/4 - 1/2 cos^-1 x`, for `- 1/sqrt2 ≤ x ≤ 1`

[Hint: Put x =  cos 2θ]

प्रमेय
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उत्तर

Put x = cos θ

∴ θ = cos–1 x

L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`

= `tan^-1 ((sqrt(1 + cos θ) - sqrt(1 - cos θ))/(sqrt(1 + cos θ) + sqrt(1 - cos θ)))`

= `tan^-1 [(sqrt(2 cos^2(θ/2)) - sqrt(2 sin^2 (θ/2)))/(sqrt(2 cos^2 (θ/2)) + sqrt(2 sin^2 (θ/2)))]`

= `tan^-1 [(sqrt(2) cos (θ/2) - sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2) + sqrt(2) sin (θ/2))]`

= `tan^-1 [((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) - (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))/((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) + (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))]`

= `tan^-1 [(1 - tan(θ/2))/(1 + tan (θ/2))]`

= `tan^-1 [(tan  pi/4 - tan (θ/2))/(1 + tan  pi/4. tan (θ/2))]  ....[∵ tan  pi/4 =1]`

= `tan^-1 [tan (pi/4 - θ/2)]`

= `pi/4 - θ/2`

= `pi/4 - 1/2 cos^-1`x  .....[∵ θ = cos–1 x]

∴ L.H.S. = R.H.S.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Inverse Trigonometric Functions - Exercise 2.3 [पृष्ठ ५२]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 2 Inverse Trigonometric Functions
Exercise 2.3 | Q 11 | पृष्ठ ५२
बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 23 | पृष्ठ ११०

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