Math, asked by ditya329, 4 days ago

solve this problem.... don't spam​

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Answered by mathdude500
3

Answer:

\qquad\boxed{ \sf{ \:\bf \:  sec\theta  + cot\theta  =  \: \dfrac{xy +  {y}^{2}  -  {x}^{2} }{x \: \sqrt{ {y}^{2}  -  {x}^{2} } }  \: }} \\  \\

Step-by-step explanation:

Given that,

\sf \: sin\theta  = \dfrac{x}{y}  \\  \\

We know,

\qquad\boxed{ \sf{ \:sin\theta  =  \frac{Opposite \: side}{Hypotenuse}  \: }} \\  \\

So, using this, we get

\sf \: \dfrac{Opposite \: side}{Hypotenuse}  = \dfrac{x}{y}  \\  \\

So, Let assume that

\sf \: Opposite \: side \:  =  \: kx \\  \\

\sf \: Hypotenuse \:  =  \: ky \\  \\

Now, By using Pythagoras Theorem, we have

\sf \:  {(Hypotenuse)}^{2} =  {(Adjacent \: side)}^{2}  +  {(Opposite \: side)}^{2}  \\  \\

\sf \:  {(ky)}^{2}  =  {(Adjacent \: side)}^{2}  +  {(kx)}^{2}  \\  \\

\sf \:  {k}^{2}{y}^{2}  =  {(Adjacent \: side)}^{2}  +   {k}^{2} {x}^{2}  \\  \\

\sf \:   {(Adjacent \: side)}^{2}  =  {k}^{2} {y}^{2} - {k}^{2} {x}^{2}  \\  \\

\sf \:   {(Adjacent \: side)}^{2}  =  {k}^{2} ({y}^{2} - {x}^{2})  \\  \\

\sf\implies  \: Adjacent \: side \:  =  \: k \sqrt{ {y}^{2}  -  {x}^{2} }  \\  \\

Now, Consider

\sf \: sec\theta  \:  +  \: cot\theta  \\  \\

\sf \:  =  \: \dfrac{Hypotenuse}{Adjacent \: side}  + \dfrac{Adjacent \: side}{Opposite \: side}  \\  \\

On substituting the values, we get

\sf \:  =  \: \dfrac{ky}{k \sqrt{ {y}^{2}  -  {x}^{2} } }  + \dfrac{k \sqrt{ {y}^{2}  -  {x}^{2} } }{kx}  \\  \\

\sf \:  =  \: \dfrac{y}{\sqrt{ {y}^{2}  -  {x}^{2} } }  + \dfrac{\sqrt{ {y}^{2}  -  {x}^{2} } }{x}  \\  \\

\sf \:  =  \: \dfrac{xy +  {y}^{2}  -  {x}^{2} }{x \: \sqrt{ {y}^{2}  -  {x}^{2} } }   \\  \\

Hence,

\bf\implies  \:  sec\theta  + cot\theta  =  \: \dfrac{xy +  {y}^{2}  -  {x}^{2} }{x \: \sqrt{ {y}^{2}  -  {x}^{2} } }   \\  \\

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