Question

Question number 26..
Please solve ASAP

Answer 1

Sol:
Given m = Cos A - Sin A    squaring on both sides we get

m2 = Cos2 A + Sin2 A - 2Sin A Cos A  = 1 - Sin 2A

And n = Cos A + Sin A    squaring on both sides we get

n2 = Cos2 A + Sin2 A + 2Sin A Cos A  = 1 + Sin 2A

⇒ (m2 + n 2) /( m2 - n2) = (1 - Sin 2A + 1 + Sin 2A ) / (1 - Sin 2A - 1 - Sin 2A )

⇒ (m2 + n 2) /( m2 - n2) =  2 /  (- 2Sin 2A )

⇒ (m2 + n 2) /( m2 - n2) =  1 /  (- 2Sin A Cos A)

⇒ (m2 + n 2) /( m2 - n2) = ( -1/2)Sec A Cosec A )

Also  (m2 + n 2) /( m2 - n2) = ( -1/2)Sec A Cosec A )

⇒ (m2 + n 2) /( m2 - n2) = ( -1/2)( 1 / Sin A Cos A )  Using identity.

⇒ (m2 + n 2) /( m2 - n2) = ( -1/2)( Sin2 A + Cos2 A ) / Sin A Cos A )

⇒ (m2 + n 2) /( m2 - n2) = ( -1/2)( Sin2 A / (Sin A Cos A ) + Cos2 A  / (Sin A Cos A )

⇒ (m2 + n 2) /( m2 - n2) = ( -1/2)( Sin A / Cos A ) + (Cos A  / Sin A )

⇒ (m2 + n 2) /( m2 - n2) = ( -1/2)( Tan A  + Cot A ).

I hope this helps u dont forget to mark this as the brainliest answer. Ask me if u have any doubts.

Answer 2

m^2=1-2sinAcosA
n^2=1+2sinAcosA

m^2+n^2 /m^2-n^2=2/-4sinAcosA=-1/2.secAcosecA

cotA+tanA=sin^2A+cos^2A/sinAcosA=secAcosecA