Question

Prove the following identity
if x= a sec 0 + b tan 0 and y = a tan 0 +b sece, prove that x2 - y2 = a²- b²​

Answer 1

Answer:

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Answer 2

$\sf{Answer}$

Given :-

x = asecθ + btanθ

y = atanθ + bsecθ

To prove :-

x² - y² = a² - b²

Solution :-

Take LHS x² - y²

x² = (asecθ + btanθ)²

y² = (a tanθ + b secθ)²

x² - y² = (asecθ + btanθ)² - (a tanθ + b secθ)²

Using ( a + b)² formula

a² + 2ab + b²

x² - y² = a² sec²θ + b²tan²θ + 2absecθtanθ - [a²tan² θ+ b²sec²θ + 2absecθtanθ]

x² - y² = a² sec²θ + b²tan²θ + 2absecθtanθ - a² tan²θ - b² sec²θ - 2ab secθ tanθ

x² - y² = a²sec²θ - a²tan²θ + b²tan²θ - b²sec²θ + 2absecθtanθ - 2absecθtanθ

x² - y² = a²sec²θ - a²tan²θ + b²tan²θ - b²sec²θ

Take common a² & b²

x² - y² = a²(sec²θ - tan²θ) + b² (tan²θ - sec²θ)

Take common "-" in (tan²θ - sec²θ)

x² - y² = a²(sec²θ - tan²θ) - b² ( -tan²θ + sec²θ)

x²-y² = a²(sec²θ - tan²θ) - b² (sec²θ - tan²θ)

We know that

sec²θ - tan²θ = 1

x² -y² = a²(1) - b²(1)

x² - y² = a² - b²

Hence proved !

LHS = RHS

Know more

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trignometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trignometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj