Question

Find the values of $\tan\bigg(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\bigg)$

Answer 1

Let $\sin^{-1}(\frac{3}{5})=x.$
then, $\sin x=\frac{3}{5}$
=>cosx=√1-sin2x=4/5 =>secx=5/4,
therefore tanx=√sec2x-1=√25/16-1=3/4
x=tan-13/4
sin-13/5=tan-13/4.....(i)
Now,cot-13/2=tan-12/3....(ii)[tan-1 1/x=cot-1x]
Hence,tan(sin-13/5+cot-13/2)
=tan(tan-13/4+tan-12/3 [using(i) and(ii)]
=tan(tan-13/4+2/-3/4.2/3)
=tan(tan-117/6)=17/6