Sin1 2x tan1 x pi2Solutionsin12x tan1x pi2 let sin12x y

Sin^-1 2x + tan^-1 x = pi/2

sin^-1(2x) + tan^-1x = pi/2

let sin^-1(2x) = y

siny = 2x

tany = 2x/sqrt(4x^2 -1)

y = tan^-1[2x/sqrt(4x^2 -1)]

So, sin^-1(2x) + tan^-1x = pi/2

tan^-1[2x/sqrt(4x^2 -1)] + tan^-1x = pi/2

we use the identity : tan^-1a +tan^-1b = tan^-1[ (a +b)/(1 -ab)]

So, tan^-1[ (a +b)/(1 -ab)] = pi/2

Now RHS = pi/2

1 -ab =0

1 = ab

2x*x/sqrt(4x^2 -1)*x =1

2x^2 = sqrt(4x^2 -1)

squaring both sides:

4x^4 = 4x^2 -1

4x^4 -4x^2 +1 =0

after solving the equation :x^2 = 1/2

x = +/- 1/sqrt2