Question

# If $$\sin { \theta } =\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } }$$, then find the values of $$\cos \theta$$ and $$\cfrac{1}{\tan { \theta } }$$

Solution

## we know that$$\sin { \theta } =\cfrac { side\quad opposite\quad to\quad angle\quad \theta }{ hypotenuse } \quad or\quad \sin { \theta } =\cfrac { Perpendicular }{ hypotenuse }$$$$\sin { \theta } =\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \Rightarrow \cfrac { P }{ H } =\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \Rightarrow \cfrac { AB }{ BC } =\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } }$$side opposite to angle $$\theta =AB={ x }^{ 2 }-{ y }^{ 2 }$$hypotenuse $$AC={ x }^{ 2 }+{ y }^{ 2 }$$In right angled $$\triangle ABC$$, we have$$\Rightarrow { \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( AC \right) }^{ 2 }\Rightarrow { \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }\Rightarrow { \left( BC \right) }^{ 2 }={ \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }-{ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }$$ (By Pythagoras theorem)$$\Rightarrow { \left( BC \right) }^{ 2 }=\left[ { x }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 } \right] \left[ { x }^{ 2 }+{ y }^{ 2 }-({ x }^{ 2 }-{ y }^{ 2 }) \right] =(2{ x }^{ 2 })(2{ y }^{ 2 })$$(using identity $${a}^{2}-{b}^{2}=(a+b)(a-b)$$)$$BC=\sqrt { 4{ x }^{ 2 }{ y }^{ 2 } } =\pm 2xy$$taking positie square root since, side cannot be negative$$\cos { \theta } =\cfrac { \quad Base }{ hypotenuse } =\cfrac { BC }{ AC } =\cfrac { 2xy }{ { x }^{ 2 }+{ y }^{ 2 } }$$and $$\tan { \theta } =\cfrac { Perpendicular }{ Base } =\cfrac { AB }{ BC } =\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 2xy }$$so, $$\cfrac { 1 }{ \tan { \theta } } =\cfrac { 1 }{ \cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 2xy } } =\cfrac { 2xy }{ { x }^{ 2 }-{ y }^{ 2 } }$$Maths

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