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Question: Prove \[{{\text{x}}^2} - {y^2} = {a^2} - {b^2}\], if \[x = a\sec \theta + b\tan \theta \] and \[y = ...

Prove x2−y2=a2−b2{{\text{x}}^2} - {y^2} = {a^2} - {b^2}, if x=asec⁡θ+btan⁡θx = a\sec \theta + b\tan \theta and y=atan⁡θ+bsec⁡θy = a\tan \theta + b\sec \theta .

Explanation

Solution

Here we will substitute the value of x and y on the left hand side of the equation and prove it equal to the right hand side using various trigonometric identities:-

1+tan⁡2θ=sec⁡2θ ⇒sec⁡2θ−tan⁡2θ=1  1 + {\tan ^2}\theta = {\sec ^2}\theta \\\ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\\

(a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab

Complete step-by-step answer:
The given equation is:-
x2−y2=a2−b2{{\text{x}}^2} - {y^2} = {a^2} - {b^2}
Let us consider the left hand side of the given equation:-
LHS=x2−y2LHS = {{\text{x}}^2} - {y^2}
Now it is given that:-
x=asec⁡θ+btan⁡θx = a\sec \theta + b\tan \theta and y=atan⁡θ+bsec⁡θy = a\tan \theta + b\sec \theta
Hence putting in the respective values of x and y in LHS we get:-
LHS=(asec⁡θ+btan⁡θ)2−(atan⁡θ+bsec⁡θ)2LHS = {\left( {a\sec \theta + b\tan \theta } \right)^2} - {\left( {a\tan \theta + b\sec \theta } \right)^2}
Now applying the following identity on both he terms:-
(a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
We get:-
LHS=(asec⁡θ)2+(btan⁡θ)2+2(asec⁡θ)(btan⁡θ)−[(atan⁡θ)2+(bsec⁡θ)2+2(bsec⁡θ)(atan⁡θ)]LHS = {\left( {a\sec \theta } \right)^2} + {\left( {b\tan \theta } \right)^2} + 2\left( {a\sec \theta } \right)\left( {b\tan \theta } \right) - \left[ {{{\left( {a\tan \theta } \right)}^2} + {{\left( {b\sec \theta } \right)}^2} + 2\left( {b\sec \theta } \right)\left( {a\tan \theta } \right)} \right]
Now simplifying it further we get:-

LHS=a2sec⁡2θ+b2tan⁡2θ+2absec⁡θtan⁡θ−[a2tan⁡2θ+b2sec⁡2θ+2absec⁡θtan⁡θ] ⇒LHS=a2sec⁡2θ+b2tan⁡2θ+2absec⁡θtan⁡θ−a2tan⁡2θ−b2sec⁡2θ−2absec⁡θtan⁡θ  LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - \left[ {{a^2}{{\tan }^2}\theta + {b^2}{{\sec }^2}\theta + 2ab\sec \theta \tan \theta } \right] \\\ \Rightarrow LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta - 2ab\sec \theta \tan \theta \\\

Now cancelling the required terms we get:-
LHS=a2sec⁡2θ+b2tan⁡2θ−a2tan⁡2θ−b2sec⁡2θLHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta
Now taking a2{a^2} and b2{b^2} common we get:-
LHS=a2(sec⁡2θ−tan⁡2θ)−b2(sec⁡2θ−tan⁡2θ)LHS = {a^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) - {b^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)
Again taking common we get:-
LHS=(sec⁡2θ−tan⁡2θ)(a2−b2)LHS = \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{a^2} - {b^2}} \right)
Now applying the following identity:-

1+tan⁡2θ=sec⁡2θ ⇒sec⁡2θ−tan⁡2θ=1  1 + {\tan ^2}\theta = {\sec ^2}\theta \\\ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\\

We get:-

⇒LHS=(a2−b2)  =RHS  \Rightarrow LHS = \left( {{a^2} - {b^2}} \right) \\\ {\text{ }} = RHS \\\

Therefore,
LHS=RHSLHS = RHS
Hence proved.

Note: The student may make mistakes while applying the identity, so the identity should be first simplified in the required form and then apply it.
The student can also use the following identity on the initial stage:
a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
Hence,
x2−y2=(x+y)(x−y){{\text{x}}^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)
Then put in the values of x and y and proceed further to get the desired answer.