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Question
Eliminate θ from the equations a sec θ – c tan θ = b and b sec θ + d tan θ = c
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Solution
Taking sec θ = x and tan θ = y we get the equations as
ax – cy = b .....(1)
bx + dy = c .....(2)
(1) × d ⇒ adx – cdy = bd ....(3)
(2) × c ⇒ bcx + cdy = c2 ......(4)
(3) + (4) ⇒ x(ad + bc) = bd + c2
∴ x = `("bd" + "c"^2)/("ad" + "bc")`
(i.e) sec θ = `("bd" "c")/("ad" "bc")`
(2) × a ⇒ ax + ay = ac ......(5)
Now (1) × b ⇒ abx – bcy = b2 ......(6)
(5) – (6) ⇒ y(ad + bc) = ac – b2
⇒ y = `("ac" - "b"^2)/("ad" + "bc")`
(i.e) tan θ = `("ac" - "b"^2)/("ad" + "bc")`
∴ sec θ = `("bd" + "c"^2)/("ad" + "bc")` and tan θ = `("ac" + "b"^2)/("ad" + "bc")`
We know sec2θ – tan2θ = 1
⇒ `(("bd" + "c"^2)^2)/(("ad" + "bc")^2) - (("ac" - "b"^2)^2)/(("ad" + "bc")^2` = 1
⇒ `(("bd" + "c"^2)^2 - ("ac" - "b"^2)^2)/(("ad" + "bc")^2` = 1
⇒ (bd+ c2)2 – (ac – b2)2 = (ad +bc)2
⇒ (bd+ c2)2 – (ad + bc)2 = (ac +b2)2
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