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Question
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
Solution
`cot(cos^-1 3/5+sin^-1x)=0`
⇒ `cos^-1 3/5+sin^-1x=cot0`
⇒ `cos^-1 3/5sin^-1x=pi/2`
⇒ `cos^-1 3/5=pi/2-sin^-1x`
⇒ `cos^-1 3/5=cos^-1x` `[becausecos^-1x=pi/2-sin^-1x]`
⇒ `x=3/5`
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