Question

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Answer 1

$\tt \:  Prove \: that \: (sec \theta - cos\theta)(cot\theta + tan\theta) = sec\theta \: tan\theta$

$\tt \:  Consider \: LHS$

$\tt \:  \: (sec \theta - cos\theta)(cot\theta + tan\theta)$

$= \tt \: (\dfrac{1}{cos\theta} - cos\theta)(\dfrac{cos\theta}{sin\theta} + \dfrac{sin\theta}{cos\theta} )$

$= \tt \:  (\dfrac{1 - {cos}^{2}\theta }{cos\theta})( \dfrac{ {cos}^{2}\theta + {sin}^{2}\theta }{sin\theta \: cos\theta} )$

$= \tt \:  (\dfrac{ {sin}^{2} \theta}{cos\theta} )(\dfrac{1}{sin\theta \: cos\theta} )$

$= \tt \:  \dfrac{sin\theta}{cos\theta \times cos\theta}$

$= \tt \:  sec\theta \: tan\theta$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1