Math, asked by NimishGulghane, 10 months ago

the answer of the attached photo with the steps​

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Answered by Anonymous
3

ANSWER:-

Given:

A + B= 90°

To prove:

 \sqrt{ \frac{tan \: A \: tan \: B + tan \: A \: cot \: B}{sin \: A \: sec \: B} -  \frac{ {sin}^{2}B }{ {cos}^{2} A}  }

Proof:

A + B = 90°

=) B= 90° - A.........(1)

Take L.H.S.

  =  >  \sqrt{ \frac{tan \: A \: tan \: B+ tan \: A \:cot \: B}{sin \: A \: sec \: B}  -  \frac{ {sin}^{2}B }{ {cos}^{2} A} }  \\  \\  =  >  \sqrt{ \frac{tan \: A\: tan(90 -A) + tan \: A \: cot \: (90 - A)}{sin \: A \: sec \: (90 - A)} -  \frac{ {sin}^{2} (90 - A)}{ {cos}^{2}A }  }  \\  \\  =  >  \sqrt{ \frac{tan \: A \: cot \: A\:  + tan \: A \:  . tan \: A}{sin \: A.cosec \: A}  -   \frac{ {cos}^{2} A }{ {cos}^{2} A }  }  \\  \\  =  >  \sqrt{ \frac{1 +  {tan}^{2} A}{1} - 1 }  \\  \\  =  >  \sqrt{ {tan}^{2}A}  \\  \\  =  > tan \: A \:  \:  \:  \: R.H.S

Hence,

Proved.

Hope it helps ☺️

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