Derivatives of inverse trigonometric functions sin-1(2x), cos-1 (x^2), tan-1 (x/2) sec-1 (1+x^2)

tan 1 by 2 ( sin inverse 2x by 1+x square + cos inverse 1-y² by 1+y² )

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

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

tan 1 by 2 (sin inverse 2x by 1+x square + cos inverse 1-y^2 by 1+y^2)

tan1/2 sin^(- 1)(2x)/(1+x^2)+cos^(- 1)(1-y^2)/(1+y^2) || (sin^ -1(2x/(1+x^2)) ||

Find the value of `tan { 1/2 sin^(-1) ((2x)/(1+x^(2))) + 1/2 cos^(-1) ((1-y^(2))/(1+ y^(2)))}`

To prove that `tan(1/2sin^-1((2x)/(1+x^2))+1/2cos^-1((1-x^2)/(1+x^2))=(2x)/(1-x^2)`

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`"tan"1/2["sin"^(-1)(2x)/(1+x^(2))+"cos"^(-1)(1-y^(2))/(1+y^(2))]` |Class 12 MATH | Doubtnut

show that (i) sin-1(2x√(1-x^2))=2sin-1 x and (ii) sin-1(2x√(1-x^2))=2cos-1 x

Q19 | Differentiate 〖tan〗^(-1)⁡(2x/(1-x^2 )) w. r. t. 〖sin〗^(-1)⁡(2x/(1+x^2 ))

Ex2.2 13. tan1/2 sin^(- 1)(2x)/(1+x^2)+cos^(- 1)(1-y^2)/(1+y^2) 12th math exercise 2.2 question13

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Trigonometry Formulas # Intermediate

Find the value of tan inverse 2 cos 2 sin inverse 1 by 2