First slide
Introduction to ITF
Question

If sin1x+sin1y+sin1z=3π2, then k=12x100k+y106kx207y207 is

Moderate
Solution

We have, sin1x+sin1y+sin1z=3π2.

It is possible only when

sin1x=π2,sin1y=π2 and sin1z=π2x=1,y=1 and z=1 k=12x100k+y106kx207y207=x100+y106+x200+y212x207y207+y207z207+z207x207=1+1+1+11+1+1=43

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