Question

11)
c) Prove that (cosec - sin e) (sec -cos ) [tan e + cot e) = 1
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Answer 1

hope this will be helpful

Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: (cosecx - sinx)(secx - cosx)(tanx + cotx) = 1$

$\large\underline{\sf{Solution-}}$

Identities Used :-

$1. \: \: \: \boxed{ \bf{cosecx = \dfrac{1}{sinx} }}$

$2. \: \: \: \boxed{ \bf{secx = \dfrac{1}{cosx} }}$

$3. \: \: \: \boxed{ \bf{tanx \: = \dfrac{sinx}{cosx} }}$

$4. \: \: \: \boxed{ \bf{cotx = \dfrac{cosx}{sinx} }}$

$5. \: \: \: \boxed{ \bf{ {sin}^{2} x + {cos}^{2} x = 1}}$

Now,

• Consider,

$\rm :\longmapsto\:(cosecx - sinx)(secx - cosx)(tanx + cotx)$

$\: \sf \: = \bigg(\dfrac{1}{sinx} - sinx \bigg) \bigg( \dfrac{1}{cosx} - cosx \bigg) \bigg( \dfrac{sinx}{cosx} + \dfrac{cosx}{sinx} \bigg)$

$\sf \: = \: \bigg(\dfrac{1 - {sin}^{2}x }{sinx} \bigg) \bigg(\dfrac{1 - {cos}^{2} x}{cosx} \bigg) \bigg( \dfrac{ {sin}^{2}x + {cos}^{2}x }{sinx \: cosx} \bigg)$

$= \: \sf \: \dfrac{ {cos}^{2} x}{sinx} \times \dfrac{ {sin}^{2}x }{cosx} \times \dfrac{1}{sinx \: cosx}$

$\sf \: = \: 1$

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

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