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Question
`sin(sin^-1 1/5+cos^-1x)=1`
Solution
`sin(sin^-1 1/5+cos^-1x)=1`
⇒ `sin^-1 1/5+cos^-1x=sin^-1 1`
⇒ `sin^-1 1/5+cos^-1x=pi/2`
⇒ `sin^-1 1/5=pi/2-cos^-1x`
⇒ `sin^-1 1/5=sin^-1x` `[thereforesin^-1x=pi/2-cos^-1x]`
⇒ `x=1/5`
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