Question

prove that [(tanθ+sinθ+1)(tanθ-secθ+1)]/tanθ=2

Answer 1

(1+cot Z-csc Z) (1+tan Z +sec Z) 

= 1 + tan Z + sec Z 

+ cot Z + 1 + cot Z*sec Z 

- csc Z - csc Z*tan Z - csc Z*sec Z 

o = opposite, a = adjacent, h = hypoteneuse 

cot Z * sin Z = (a/o)*(h/a) = h/o = csc Z 

csc Z*tan Z = (h/o)*(o/a) = h/a = sec Z 

csc Z*sec Z = (h/o)*(h/a) 


= 1 + tan Z + sec Z 

+ cot Z + 1 + csc Z 

- csc Z - sec Z - csc Z*sec Z 

= 1 + tan Z 

+ cot Z + 1 

- csc Z*sec Z 

= 2 + tan Z + cot Z - csc Z*sec Z 

= 2 + o/a + a/o - (h/o)*(h/a) 

= 2 + (o^2 + a^2 - h^2)/(a*o) 

Pythagoras... 

= 2 <<<<<.......

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Answer 2

(1+cotθ-cosecθ) (1+tanθ +secθ) 
= 1 + tanθ + secθ + cotθ + cotθtanθ + cotθsecθ - cosecθ - tanθcosecθ - secθcosecθ 
Now we know that cotθtanθ = 1 and cotθsecθ = cosecθ and tanθcosecθ = secθ 

= 1 + tanθ + secθ + cotθ + 1 + cosecθ - cosecθ - secθ - secθcosecθ 
= 1+ tanθ + cotθ + 1 - secθcosecθ 
= 2 + [tanθ + cotθ - secθcosecθ] 
= 2 + [sinθ/cosθ + cosθ/sinθ - secθcosecθ] 
= 2+ [(sin^2θ + cos^2θ)/sinθcosθ - secθcosecθ] 

Or 
(1+cotθ-cosecθ) (1+tanθ +secθ) 
= (1+ cosθ/sinθ - 1/sinθ) (1+ sinθ/cosθ + 1/cosθ) 
= 1/sinθ[sinθ + cosθ -1] * 1/cosθ[sinθ + cosθ +1] 
= 1/sinθcosθ [(sinθ + cosθ)^2 - 1] 
= 1/sinθcosθ[sin^2θ + 2sinθcosθ + cos^2θ -1] 
Using the identity sin^2θ + cos^2θ = 1 
= 1/sinθcosθ[1 + 2sinθcosθ -1] 
= 2sinθcosθ/sinθcosθ 
= 2 

Now using the identity sin^2θ + cos^2θ =1 
= 2+ [1/sinθcosθ - secθcosecθ] 
= 2 + secθcosecθ - secθcosecθ 
=2

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