Advertisements
Advertisements
Question
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`
Solution
We have to find the value of `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))]`
We know that: `sin^(-1)(2x)/(1+x^2)=2tan^(-1)x for |x| ≤ 1 …… (1)`
`cos^(-1)(1-y^2)/(1+y^2)=2tan^(-1)y for y > 0 …… (2)`
`Now sin^(-1)((2x)/(1+x^2)) + cos^(-1)((1-y^2)/(1+y^2))=2tan^(-1)x+2tan^(-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)`
Since, ` tan^(−1)x + tan^(−1)y = tan^(−1)((x+y)/(1-xy)) for xy < 1`
`therefore tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))]=tan(tan^(−1)((x+y)/(1-xy)))=(x+y)/(1-xy)`
APPEARS IN
RELATED QUESTIONS
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
`sin^-1(sin4)`
Evaluate the following:
`tan^-1(tan12)`
Evaluate the following:
`sec^-1(sec (25pi)/6)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`cot^-1(cot (9pi)/4)`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`sin(sec^-1 17/8)`
Evaluate the following:
`cosec(cos^-1 3/5)`
Evaluate the following:
`cot(cos^-1 3/5)`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Prove the following result
`sin(cos^-1 3/5+sin^-1 5/13)=63/65`
Solve: `cos(sin^-1x)=1/6`
Evaluate:
`cos(tan^-1 3/4)`
`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3`
Find the value of the following:
`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1
Prove that `2tan^-1(sqrt((a-b)/(a+b))tan theta/2)=cos^-1((a costheta+b)/(a+b costheta))`
For any a, b, x, y > 0, prove that:
`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1 (2alphabeta)/(alpha^2-beta^2)`
`where alpha =-ax+by, beta=bx+ay`
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]
If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\] = α, then x2 =
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
If sin−1 x − cos−1 x = `pi/6` , then x =
The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is
If \[\cos^{- 1} x > \sin^{- 1} x\], then
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`
