Math, asked by Neervineeti, 4 months ago

If sin θ =c/√c^2+d^2,
where d >0 then find the values of cos θ and tan θ. ​

Answers

Answered by anindyaadhikari13
1

SOLUTION:-

Given that,

 \sf \sin(x)  =  \frac{c}{ \sqrt{ {c}^{2} +  {d}^{2}  } }

We have to find the value of cos(x) and tan(x)

We know that,

 \sf { \sin}^{2} (x) +  { \cos}^{2} (x) = 1

 \sf \implies{ \cos} (x) =  \sqrt{1 -  { \sin }^{2}(x) }

 \sf =  \sqrt{1 -  \frac{ {c}^{2} }{ {c}^{2} +  {d}^{2}  } }

 \sf =  \sqrt{  \frac{ {c}^{2} +  {d}^{2}   -  {c}^{2} }{ {c}^{2} +  {d}^{2}  } }

 \sf =  \sqrt{ \frac{ {d}^{2} }{ {c}^{2}  +  {d}^{2} } }

 \sf =  \frac{d}{ {c}^{2} +  {d}^{2}  }

Hence,

 \sf \implies \cos(x) =   \frac{d}{ {c}^{2} +  {d}^{2}  }

Now,

 \sf \tan(x) =  \frac{ \sin(x) }{ \cos(x) }

 \sf =  \frac{c}{ {c}^{2} +  {d}^{2} }  \div  \frac{d}{ {c}^{2} +  {d}^{2} }

 \sf =  \frac{c}{ \cancel{ {c}^{2} +  {d}^{2}} } \times  \frac{ \cancel{{c}^{2} +  {d}^{2}}}{d}

 \sf =  \frac{c}{d}

Answer:-

  1.  \sf  \cos(x) =   \frac{d}{ {c}^{2} +  {d}^{2}  }
  2.  \sf  \tan(x) =  \frac{c}{d}
Answered by nehashanbhag0729
0

Answer:

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