Math, asked by suraj5912, 10 months ago

root under sec square A + cosec square A = secA . cosec A​

Answers

Answered by drjiya123
0

under root sec square A =secA

under root cosec square A=CosecA

=SecA.CosecA=RHS

Hope it helps

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Answered by Anonymous
0

Solution

 \sqrt{ {sec}^{2} \: \theta +  cosec^{2} \: \theta  }  =  \sec \:\theta \times cosec \: \: \theta

LHS

 =  \sqrt{ {sec}^{2} \: \theta +  cosec^{2} \: \theta  }

we know that →

  •  sec^{2}\theta =  \frac{1}{ {cos}^{2}\theta }
  •  cosec^{2}\theta =  \frac{1}{ {sin}^{2}\theta }

Putting tha value of

 {cosec}^{2} \theta and {sec^{2}\theta}

 \sqrt{ \frac{1}{ {cos}^{2}\theta } +  \frac{1}{ {sin}^{2}\theta }  }

 =  \sqrt{ \frac{ {sin}^{2}\theta +  {cos}^{2}\theta  }{ {cos}^{2}\theta \times  {sin}^{2}\theta  } }

We know that →

  •  {sin}^{2} \theta +  {cos}^{2} \theta = 1

Then,

 =  \sqrt{ \frac{1}{ {cos}^{2}\theta \times  {sin}^{2}\theta  } }

 =  \frac{1}{sin \: \theta \times cos\theta}

We can write it as ,

 = cosec \: \theta \times  \sec \: \theta

\boxed{=sec \: \theta × cosec\: \theta} RHS

Hence Proved !!

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