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### Trigonometric identities - 15

Trigonometric identity: (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = 2

Taking the left hand side expression of the given identity,
(1 + cot θ - cosec θ)(1 + tan θ + sec θ)
Applying the following identities
• cot θ = cos θ / sin θ
• cosec θ = 1 / sin θ
• tan θ = sin θ / cos θ
• sec θ = 1 / cos θ
(1 +  [cos θ / sin θ] -  [1 / sin θ ])(1 +  [sin θ / cos θ] +  [1 / cos θ])
Simplifying the fractions by taking a common denominator,
([sin θ + cos θ - 1] / sin θ)([cos θ + sin θ + 1] / cos θ)
Multiplying the two rational expressions by multiplying together their numerators and denominator,
[(sin θ + cos θ - 1)(cos θ + sin θ + 1)] / [(sin θ)(cos θ)]
Let a = sin θ + cos θ, then the fraction becomes
[(a - 1)(a + 1)] / [(sin θ)(cos θ)]
Applying identity a2 - b2 = (a + b)(a - b),
[a2 - 12] / [(sin θ)(cos θ)]
Put back a = sin θ + cos θ,
[(sin θ + cos θ)2 - 12] / [(sin θ)(cos θ)]
Simplifying the numerator,
[sin2θ + cos2θ + 2(sinθ)(cosθ) - 1] / [(sin θ)(cos θ)]
Applying identity sin2θ + cos2θ = 1,
[1 +  2(sinθ)(cosθ) - 1] / [(sin θ)(cos θ)]
Simplifying the numerator,
[2(sinθ)(cosθ)] / [(sinθ)(cosθ)]
Cancelling common (sinθ)(cosθ) from the numerator and denominator,
= 2
... which is the RHS expression.

Identities applied:
• cot θ = cos θ / sin θ
• cosec θ = 1 / sin θ
• tan θ = sin θ / cos θ
• sec θ = 1 / cos θ
• sin2θ + cos2θ = 1