Suppose Sax=Sec-1xa+Sec-1a for a≠0. If Sa(x)=Sb(x) for a≠b then x=

sa(x)=sb(x) sec-1xa+sec-1a=sec-1xb+sec-1b sec-1xa-sec-1xb=sec-1(b)-sec-1a xa=b ⇒ x=ab

Suppose Sax=Sec-1xa+Sec-1a for a≠0. If Sa(x)=Sb(x) for a≠b then x=

sa(x)=sb(x) sec-1xa+sec-1a=sec-1xb+sec-1b sec-1xa-sec-1xb=sec-1(b)-sec-1a xa=b ⇒ x=ab

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Suppose $S_{a} (x) = {Sec}^{- 1} (\frac{x}{a}) + {Sec}^{- 1} (a)$ for $a \neq 0 .$ If $S_{a} (x) = S_{b} (x)$ for $a \neq b$ then $x =$

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$s_{a} (x) = s_{b} (x) \sec^{- 1} (\frac{x}{a}) + \sec^{- 1} a = \sec^{- 1} (\frac{x}{b}) + \sec^{- 1} b \sec^{- 1} (\frac{x}{a}) - \sec^{- 1} (\frac{x}{b}) = \sec^{- 1} (b) - \sec^{- 1} a \frac{x}{a} = b \Rightarrow x = ab$

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Suppose Sax=Sec-1xa+Sec-1a for a≠0. If Sa(x)=Sb(x) for a≠b then x=

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Suppose $S_{a} (x) = {Sec}^{- 1} (\frac{x}{a}) + {Sec}^{- 1} (a)$ for $a \neq 0 .$ If $S_{a} (x) = S_{b} (x)$ for $a \neq b$ then $x =$