Prove that:

(1) sin²/ cos + cos = (sin²+cos²)/cos
= 1/cos [sin²+cos² = 1]
= sec [1/cos = sec]
Hence proved.

(2) cos² (1 + tan² ) = cos² +sin² [cos² ×tan² = cos² ×sin²/cos² = sin²]
= 1 [sin²+cos² = 1]
Hence proved.

(3) √[(1-sin)/(1+ sin)] = √[(1-sin)/(1+ sin)]×√[(1- sin)/(1- sin)] [rationalizing denominator]
= √[(1-sin)²/(1-sin²)]
= √[(1-sin)²/cos² [1-sin² = cos² ]
= (1-sin)/cos [taking square root]
= (1/cos)-(sin/cos)
= sec-tan [1/cos = sec , sin/cos = tan
Hence proved.

(4) ( sec – cos)( cot + tan) = LHS
∵sec = 1/cos , cot = cos/sin , tan = sin/cos
LHS = (1/cos)-cos][(cos/sin)+(sin/cos)]
LHS = [(1-cos²)/cos][( cos²+sin²)/(soncos)]
= [sin²/cos][1/sincos] [∵1-cos² = sin²]
= sin/cos²
= sectan [sin/cos = tan, 1/cos =sec]
= RHS
Hence proved.

(5) cot + tan = (cos/sin) +(sin/cos) [∵cot = cos/sin , tan = sin/cos]
= (cos²+ sin²)/sincos
= 1/ sincos [∵cos²+ sin²= 1]
= cosecsec [1/ sin = cosec , 1/cos = sec]
Hence proved.

(6) 1/(secθ-tanθ) = 1/(secθ-tanθ) × (sec + tan)/(sec+tan) [rationalising denominator]
= (sec+tan)/( sec²-tan²)
= sec+tan [∵sec²-tan² = 1]
Hence proved.