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Mathematics Question on Inverse Trigonometric Functions

If sin⁡(sin−115+cos⁡−1x)=1\sin(sin^{-1}\frac{1}{5}+\cos^{-1}x)=1 , then find the value of x.

Answer

sin⁡(sin−115+cos⁡−1x)=1\sin(sin^{-1}\frac{1}{5}+\cos^{-1}x)=1
⇒sin⁡(sin⁡−115)cos⁡(cos⁡−1)+cos⁡(sin⁡−115)sin⁡(cos⁡−1x)=1\Rightarrow\sin(\sin^{-1}\frac{1}{5})\cos(\cos^{-1})+\cos(\sin^{-1}\frac{1}{5})\sin(\cos^{-1}x)=1
[sin(A+B)=sinAcosB+cosAsinB]
⇒15∗x+cos⁡(sin⁡−115)sin⁡(cos⁡−1x)=1\Rightarrow\frac{1}{5}*x+\cos(\sin^{-1}\frac{1}{5})\sin(\cos^{-1}x)=1
⇒15+cos⁡(sin⁡−115)sin⁡(cos⁡−1x)=1\Rightarrow\frac{1}{5}+\cos(\sin^{-1}\frac{1}{5})\sin(\cos^{-1}x)=1 .......(1)
Now, let sin-1(15)(\frac{1}{5})=y.
Then ,sin y=15\frac{1}{5} ⇒\Rightarrow cos y= 1−(152)=265\sqrt{1-(\frac{1}{5}^2})=2\sqrt{\frac{6}{5}}
⇒\Rightarrow cos⁡−1(265)\cos^{-1}(2\frac{\sqrt6}{5}).
therefore sin-11/5=cos-1(2√6/5) .......(2)
let cos-1 x=z
Then cosz=x =>sinz=√1-x2
=>z=sin-1(√1-x2)
therefore cos-1x=sin-1(√1-x2) ....(3)
from (1),(2) and (3)
x/5+cos(cos-1(2√6/5)).sin(sin-1(√1-x2))=1
x/5+2√6/5.√1-x2=1
⇒\Rightarrow x+2√6 1−x2\sqrt{1-x^2} =5
⇒\Rightarrow 2√6 1−x2\sqrt{1-x^2} =5-x
On squaring both sides, we get:(4)(6)(1-x2)=25+x2-10x
⇒\Rightarrow 24-24x2=25+x2-10x
⇒\Rightarrow 25x2-10x+1=0
⇒\Rightarrow (5x-1)2=0
⇒\Rightarrow x=15\frac{1}{5}.

Hence, the value of x is 15\frac{1}{5}