Question

plzz ans d 3rd one plzzzzzzzz​

Answer 1

$\huge \bold{QUESTION \: 3}$

Given,

$\mathsf{ m = a \: sec \: A + b \: tan \: A}$

$\mathsf{ n= a \: tan \: A + b\: sec\: A }$

$\huge \boxed{\bold{\bigstar {m}^{2} - {n}^{2} = {a}^{2} - {b}^{2}}}$

On taking Left hand side (L. H. S)

$\mathsf{ {m}^{2} - {n}^{2} = ( a \: sec \: A + b \: tan \: A)^{2} - (a \: tan \: A + b \: sec \: A )^{2}}$

By identity ,

$\huge\boxed{(a + b)^{2}= {a}^{2} +{b}^{2} + 2ab}$

$\mathsf{ {m}^{2} - {n}^{2} = {a}^{2} sec^{2} A + {b}^{2} tan^{2} A + 2 ab \: sec A \:tan A - ({a}^{2} tan^{2} A + {b}^{2} sec^{2} A + 2 ab \: tan\: A\: sec\: A)}$

$\mathsf{ {m}^{2} - {n}^{2} = {a}^{2} sec^{2} A + {b}^{2} tan^{2} A + \cancel{2 ab \: sec\: A \: tan \: A }- ({a}^{2} tan^{2} A - {b}^{2} sec^{2} A + \cancel{2 ab \: sec\: A \: tan \: A })}$

$\mathsf{ sec^{2} A (a^{2} - b^{2}) - </p><p>tan^{2} A (a^{2} - b^{2}) }$

$\mathsf{ (a^{2} - b^{2}) (sec^{2} A -</p><p>tan^{2} A )}$

Since,

$\huge \boxed{(sec^{2} A -</p><p>tan^{2} A ) = 1}$

$\mathsf{ (a^{2} - b^{2}) = R. H. S}$

$\large \mathcal{HENCE \: PROVED}$