Evaluate tan sin1 23 sin cos1 38SolutionA tansin123 Let sin1

Evaluate: tan (sin^-1 2/3) sin (cos^-1 (3/8))

Solution

A tan(sin^-12/3)

Let sin^-12/3=t

Therefore sin t=2/3

Here opposite=2 , hypotenuse=3

Adjacent= sqrt(3²-2²)= sqrt 5

tan t= opposite/adjacent = 2/ sqrt5 = (2sqrt5)/5

Therefore tan(sin^-12/3)= (2sqrt5)/5

B. Sin(cos^-13/8)

Cos^-13/8=t

Cos t=3/8

Here adjacent=3 , hypotenuse=8

Opposite=sqrt(8²-3²) = sqrt 75

Sin t= opposite/hypotenuse = sqrt75 /8

Hence sin(cos^-13/8) = sqrt75 /8