Question

solve THIS 50 points maths

Answer 1

Answer:

QED

Step-by-step explanation:

replacing thetha by x make easy in typing

LHS =

(sinx + secx )^2 + (cosx+Cosecx)^2

= (Sinx + I/cosx)^2 + (cosx + 1/sinx)^2

= Sin^2x + 1/Cos^2x + 2Sinx/Cosx + Cos^2x + 1/Sin^2x + 2Cox/Sinx

= Sin^2x + Cos^2x  + (1/Cos^2xSin^2x)(Sin^2x+2Sin^3xCosx+Cos^2x+2Cos^3xSinx)

= 1 + (1/Cos^2xSin^2x)(Sin^2x + Cos^2x  + 2sinxCox(Sin^2x + Cos^2x))

= 1 + (1/Cos^2xSIn^2x)(1+2sinxCosx)

= 1 + 1/Cos^2xSin^2x + 2/cosxSinx

RHS

= (1+SecxCosecx)^2

= (1 + 1/cosxSinx)^2

= 1 + 1/cos^2xSin^2x + 2/CosxSinx

so LHS = RHS

QED

Answer 2

Answer:

replacing thetha by x make easy in typing

LHS =

(sinx + secx ^2+ (cosx+Cosecx)^2

= (Sinx + 1/cosx)^2 + (cosx + 1/sinx)^2 Sin^2x + 1/Cos^2x + 2Sinx/Cosx +

Cos 2x + 1/Sin 2x + 2Cox/Sinx

= Sin^2x + Cos"2x + (1/Cos 2xSin^2x)(Sin ^2x+2Sin^3xCosx+Cos^2x+2Cos^3xSinx)

= 1 + (1/Cos2xSin^2xX(Sin^2x + cos2x + 2sinxCox(Sin^2x + Cos^2x))

= 1 + (1/Cos 2xSln^2x)(1+2sinxCosx)

= 1 + 1/Cos 2xSin^2x + 2/cosxSinx

RHS

= (1+SecxCosecx)^2

= (1 + 1/cosxSinx)^2

= 1+ 1/cos^2xSin^2x + 2/CosxSinx

so LHS = RHS

QED