Question

# Inverse circular functions,Principal values of $${ sin }^{ -1 }x,{ cos }^{ -1 }x,{ tan }^{ -1 }x$$.$${ tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy }$$,      $$xy<1$$       $$\pi +{ tan }^{ -1 }\frac { x+y }{ 1-xy }$$,         $$xy>1$$.(a) $${ cot }^{ -1 }9+{ cosec }^{ 1 }\frac { \sqrt { 41 } }{ 4 } =\frac { \pi }{ 4 }$$(b) $${ tan }^{ -1 }\frac { 1 }{ 7 } +{ tan }^{ -1 }\frac { 1 }{ 13 } ={ tan }^{ -1 }\frac { 2 }{ 9 }$$(c) $${ tan }^{ -1 }2+{ tan }^{ -1 }3=3\pi /4$$

Solution

## (a) $${ cosec }^{ -1 }x={ cot }^{ -1 }\sqrt { { x }^{ 2 }-1 }$$L.H.S.$$={ cot }^{ -1 }9+{ cot }^{ -1 }\sqrt { (41/16)-1 }$$         $$={ cot }^{ -1 }9+{ cot }^{ -1 }(5/4)$$        $$={ tan }^{ -1 }\dfrac { 1 }{ 9 } +{ tan }^{ -1 }\dfrac { 4 }{ 5 }$$       $$={ tan }^{ -1 }\dfrac { (1/9)+(4/5) }{ 1-(1/9).(4/5) } ={ tan }^{ -1 }\dfrac { 41 }{ 41 }$$       $$={ tan }^{ -1 }1=\pi /4$$(b) Do yourself(c) $${ tan }^{ -1 }2+{ tan }^{ -1 }3=\pi +{ tan }^{ -1 }\dfrac { 2+3 }{ 1-2.3 } ,$$$$\because xy=2.3=6>1$$     $$=\pi +{ tan }^{ -1 }\{ 5/(-5)\}$$     $$=\pi +{ tan }^{ -1 }(-1)=\pi -{ tan }^{ -1 }1$$     $$\pi -\pi /4=3\pi /4$$Mathematics

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