Question

sin theta + sec theta whole square + cos theta + cosec theta whole square is equal to 1 + sec theta cosec theta whole square

Answer 1

Answer:

LHS=RHS

Step-by-step explanation:

Refer to attachment

Answer 2

Answer:

Consider angle theta as A

$\longrightarrow$ (sinA + Sec A)² + (cos A + cosec A)² 

$\longrightarrow$  [ (SinA CosA + 1)²/CosA ]² + [ (CosA SinA + 1)² / Sin²A ]

$\longrightarrow$  (1+ SinA CosA)² * [1 / Cos²A  + 1/sin²A ]

$\longrightarrow$ (1 + SinA CosA)² * [ Cos²A + Sin²A]/ [cos²A * Sin²A ]

$\longrightarrow$ ( 1 + SinA CosA)²/ (cosA * SinA)²

$\longrightarrow$ [ 1/cosA * 1/sinA  + 1 ] ²

 = [ Sec A Cosec A + 1 ]²