If A and B are acute angle of right angled Triangle ABC the prove TanA.tanB is equal to 1

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1. [tex]\large\underline{\bf{Solution-}}[/tex]

   Given :-

   In triangle ABC,

   • A and B are acute angles and triangle is right angled at C.

   To Prove :-

   • tanA × tanB = 1

   Solution :-

   Given that,

   • Triangle ABC is right-angle triangle right-angled at C.

   So,

   ⟼ Using angle sum property of triangle,

   ⟼ ∠A + ∠B + ∠C = 180°.

   ⟼ ∠A + ∠B = 180° – ∠C

   ⟼ ∠A + ∠B = 180° – 90°

   ⟼ ∠A + ∠B = 90°

   ⇛ ∠B = 90° – ∠A

   Consider,

   ⟼ tanA × tanB

   = tanA × tan(90° – A)

   = tanA × cotA

   [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: tan(90 \degree \: – x) = cotx\bigg \}}[/tex]

   = 1

   [tex]\red{\bigg \{ \because \: tanx = \dfrac{1}{cotx}\rm :\implies\:tanx \times cotx = 1 \bigg \}}[/tex]

   [tex]\large{\boxed{\boxed{\bf{Hence, Proved}}}}[/tex]

   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

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