Question

only answer the 7th question....fast....​

Answer 1

Answer:

It's easy.

Step-by-step explanation:

Rearrange the given equations;

• $\frac{x}{p} =sec$θ.....(1)

• $\frac{y}{q} = tan$θ.....(2)

Square both equations and subtract them

L.H.S= $\frac{x^2}{p^2} - \frac{y^2}{q^2}$ ⇒ $\frac{x^2q^2-y^2p^2}{p^2q^2}$

R.H.S= $sec^2$θ $- tan^2$θ = 1 {from identity}

⇒ $x^2q^2-y^2p^2=p^2q^2$

Hence proved

Hope this is clear

Have a good day!

Answer 2

Answer:

x = p secθ + q tanθ and y = p tanθ + q secθ

L.H.S. = x2 - y2 

= (p secθ + q tanθ)2 - (p tanθ + q secθ)2

= p2 sec2θ + 2pq secθ tanθ + q2 tan2θ - (p2tan2θ + 2pq tanθ secθ + q2sec2θ)

= p2sec2θ + 2pq secθ tanθ + q2 tan2θ - p2 tan2θ - 2pq tanθ secθ - q2 sec2θ

= (p2-q2) sec2θ + (q2-p2) tan2θ

= (p2-q2) sec2θ + (q2-p2) tan2θ = (p2-q2) (sec2θ - tan2θ)

= (p2-q2) [since 1 + tan2θ = sec2θ]

= R.H.S. ∴ x2-y2 = p2-q2.