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Find the value of the following: tan 1/2[sin-1 2x/1+ x2 + cos-1 1-y2/1+y2], |x| < 1, y > 0 and xy < 1 - Mathematics

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प्रश्न

Find the value of the following:

`tan  1/2 [sin^(-1)  (2x)/(1+ x^2) + cos^(-1)  (1-y^2)/(1+y^2)]`, |x| < 1, y > 0 and xy < 1

योग
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उत्तर

Let x = tan θ

Then, θ = tan−1 x

∴ `sin^(-1)  (2x)/(1+x^2 ) `

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

= `sin^(-1) (sin 2 θ)`

= 2θ

= 2 tan−1 x

Let y = tan `phi`

Then, `phi` = tan−1 y

∴ `cos^(-1)  (1 - y^2)/(1+ y^2)`

= `cos^(-1) ((1 - tan^2 phi)/(1+tan^2 phi))`

= `cos^(-1)(cos 2phi)`

= `2phi`

= 2 tan−1 y

∴ `tan  1/2 [sin^(-1)  (2x)/(1+ x^2) + cos^(-1)  (1-y^2)/(1+y^2)]`

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

= `tan[tan^(-1) x + tan^(-1) y]`

= `tan[tan^(-1) ((x+y)/(1-xy))]`

= `(x+y)/(1-xy)`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Inverse Trigonometric Functions - Exercise 2.2 [पृष्ठ ४८]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 2 Inverse Trigonometric Functions
Exercise 2.2 | Q 13 | पृष्ठ ४८

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