Question

please can you give solution of this​

Answer 1

Answer:

1

Step-by-step explanation:

x=a sin teta

x/a=sin teta

a/x=1/sin teta=cosec teta

y=b tan teta

y/b=tan teta

b/y=1/tan teta=cot teta

(a/x)^2-(b/y)^2=cosec^2 teta-cot^2 teta

=1(formula)

Answer 2

Given :-

x = a sin θ

y = b tan θ

To find :-

(a²/x²)-(b²/y²) = 1

Solution :-

Given that

x = a sin θ

=> x/a = sin θ

=> 1/(x/a) = 1/ sin θ

=> a/x = cosec θ

On squaring both sides then

(a/x)² = (cosec θ)²

=> a²/x² = cosec² θ ------------(1)

and

Given that

y = b tan θ

=> y/b = tan θ

=> 1/(y/b) = 1/ tan θ

=> b/y = cot θ

On squaring both sides then

=> (b/y)² = ( cot θ)²

=> b²/y² = cot² θ -------------(2)

On subtracting (2) from (1)

=> (1) - (2)

=> (a²/x²)-(b²/y²) = cosec² θ - cot² θ

Therefore , (a²/x²)-(b²/y²) = 1

Since, cosec² A - cot² A = 1

Hence, Proved.

Answer :-

(a²/x²)-(b²/y²) = 1

Used formulae:-

• cosec θ = 1/ sin θ

• cot θ = 1/ tan θ

• cosec² A - cot² A = 1