Question

Prove that (sinA+cosecA)^2+(cosA+secA)^2 equal to 7+tan^2A+cot^2A​

Answer 1

Answer:

sin^2A+cosec^2A+2sinAcosecA+cos^2A+sec^2A+2cosAsecA

(sin^2A+cos^2A)+2+2+sec^2A+cosec^2A

1+4+sec^2A+cosec^2A

5+1+tan^2A+1+cot^2A

7+tan^2A+cot^2A

Answer 2

TO PROVE THAT :-

$\sf (sinA+cosecA)^2+(cosA+secA)^2= 7+tan^2A+cot^2A$

SOLUTION :-

$\sf L.H.S = (sinA+cosecA)^2+(cosA+secA)^2$

         $= \sf sin^2A+cosec^2A+2sinAcosecA+cos^2A+sec^2A+2cosAsecA$

$\bullet\bf \ sin^2A+cos^2= 1$

        $= \sf 1+cosec^2A+sec^2A+2sinAcosecA+2cosAsecA$

$\bullet\bf \ sinA=\dfrac{1}{cosecA}\\\\\bullet \ cosA= \dfrac{1}{secA}$

       $= \sf 1+cosec^2A+sec^2A+ \left( 2\times \dfrac{1}{cosecA}\times cosecA\right)+\left(2\times \dfrac{1}{secA}\times secA \right) \\\\= 1+2+2+cosec^2A+sec^2A \\\\= 5+cosec^2A+sec^2A$

$\bullet\bf \ sec^2A= 1+tan^2A\\\\\bullet \ cosec^2A= 1+cot^2A$

       $= \sf 5+(1+cot^2A)+(1+tan^2A)\\\\= 5+1+1+tan^2A+cot^2A\\\\= 7+tan^2A+cot^2A\\\\= R.H.S$

Hence proved.