Question

(sinx+cosecx)^2+(cosx+secx)^2=tan^2x+cot^2x+7​

Answer 1

Answer:

Step by step explanation :

Let us take the LHS and find and let the RHS be as it is

(Sinx+cosecx)^2+(cosx+secx)^2

We know the equation (a+b)^2=a^2+b^2+2ab;put the equation to both of the above ;

=(Sin^2x + cosec^2x +2sinx cosecx) +(cos^2x +sec^2x+2cosxsecx)

=(sin^2x+cosec^2x+2sinx 1/sinx) +(cos^2x+sec^2x+2cosx 1/cosx)

= (sin^2x + 1+cot^2x +2) + (cos^2x+1+tan^2x+2)

= (sin^2x +cot^2x+3)+(cos^2x+tan^2x+3)

=sin^2x+cos^2x +cot^2x +tan^2x +6

=1 + cot^2x +tan^2x +6

=tan^2x +cot^2x +7

Hence the LHS =RHS