Question
If $${\sin ^{ - 1}}\left( {\frac{x}{5}} \right) + {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2},$$ then the value of $$x$$ is
A.
4
B.
5
C.
1
D.
3
Answer :
3
Solution :
$$\eqalign{
& {\sin ^{ - 1}}\left( {\frac{x}{5}} \right) + {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2} \cr
& \Rightarrow \,\,{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) = \frac{\pi }{2} - {\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{5}{4}} \right) \cr
& \Rightarrow \,\,{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) \cr
& \left[ {\because \,\,{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}} \right] \cr
& \Rightarrow \,\,{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) = {\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{4}{5}} \right) \cr
& \Rightarrow \,\,{\sin ^{ - 1}}\frac{x}{5} = {\sin ^{ - 1}}\frac{3}{5} \cr
& \Rightarrow \,\,\frac{x}{5} = \frac{3}{5} \cr
& \Rightarrow \,\,x = 3 \cr} $$