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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Show that $cot^{-1}7+cos^{-1}8+cot^{-1}18=cot^{-1}3$

$(a)\;cos^{-1}3\qquad(b)\;cot^{-1}3\qquad(c)\;sin^{-1}3\qquad(d)\;tan^{-1}3$

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1 Answer

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Toolbox:
  • $cot^{-1}x=tan^{-1}\large\frac{1}{x}$
  • $tan^{-1}x+tan^{-1}y=tan^{-1}\bigg(\large\frac{x+y}{1-xy}\bigg)$
L.H.S.=
$cot^{-1}7+cos^{-1}8+cot^{-1}18=tan^{-1}1/7+tan^{-1}1/8+tan^{-1}1/18$
$tan^{-1}1/7+tan^{-1}1/8=tan^{-1}\bigg(\large\frac{1/7+1/8}{1-1/7.1/8}\bigg)$
$=tan^{-1}\bigg(\large\frac{15/56}{55/56}\bigg)$$=tan^{-1}\large\frac{15}{55}$$=tan^{-1}\large\frac{3}{11}$
$\Rightarrow\:tan^{-1}1/7+tan^{-1}1/8+tan^{-1}1/18=$
$tan^{-1}3/11+tan^{-1}1/18=tan^{-1}\bigg(\large\frac{3/11+1/18}{1-3/11.1/18}\bigg)$
$=tan^{-1}\bigg(\large\frac{65/198}{195/198}\bigg)$
$=tan^{-1}\large\frac{1}{3}$$=cot^{-1}3$=R.H.S
answered Mar 21, 2014 by rvidyagovindarajan_1
 

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