If x=p.sec and y=q.tan, then choose the correct equation.
a)x2-y2=p2q2.
b)x2q2-y2p2=pq.
c)x2q2-y2p2=1/(p2q2).
d)x2q2-y2p2=p2q2.
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If x=p.sec and y=q.tan, then choose the correct equation.
a)x2-y2=p2q2.
b)x2q2-y2p2=pq.
c)x2q2-y2p2=1/(p2q2).
d)x2q2-y2p2=p2q2.
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Step-by-step explanation:
Solution:-
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.
hope this helps you ☺️
Answer:
option d x²q² - y²p² = p²q²
Step-by-step explanation: This is like the more easy method
So we know that, Sec²- tan² = 1
Substituting Sec theeta = x/p and tan theeta = y/q
so we have (x/p)² - (y/q)² = 1
and we make the denominator same by, x²×q²/p²×q² - y²×p²/q²×p² = 1
x²q² - y²p²/p²q² = 1
we bring the denominator to the LHS and we get,
x²q² - y²p² = 1×p²q²
x²q² - y²p² = p²q²
Hope it helps!