Question

prove that tanA/cotA= sec^2A/cosec^2A​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

GIVEN TO PROVE

$\dfrac{tanA}{cotA} = \dfrac{sec^2A}{cosec^2A}$

TO KNOW :-

tanA = sinA/cosA

cotA = cosA/sinA

secA = 1/cosA or cosA = 1/secA

cscA = 1/sinA or sinA = 1/cscA

PROOF !

Take LHS

$\dfrac{tanA}{cotA}$

$\dfrac{sinA/cosA}{cosA/sinA}$

$\dfrac{sinA\times sinA}{cosA \times cosA}$

$\dfrac{sin^2A}{cos^2A}$

$\dfrac{1/csc^2A}{1/sec^2A}$

$\dfrac{sec^2A}{csc^2A}$

Hence ,

LHS = RHS

PROVED!

KNOW MORE :-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj