Question

If $ x = a\sec \theta + b\tan \theta $ and $ y = a\tan \theta + b\sec \theta $ , prove that $ {x^2} - {y^2} = {a^2} - {b^2} $ .

Answer 1

Hint: We know that the above question is connected to the trigonometry as we know that $ \sec ,\tan $ are the trigonometric ratios. Here we will take the given value and then solve the left hand side of the equation by using trigonometric identities. We will also use some of the basic sum of squares formula like $ {(a + b)^2} = {a^2} + 2ab + {b^2} $ .

Complete step by step solution:
As per the given question we have $ x = a\sec \theta + b\tan \theta $ and $ y = a\tan \theta + b\sec \theta $ . We will take the left hand side of the equation i.e. $ {x^2} - {y^2} $ . Now substituting the value we have,
 $ \Rightarrow {x^2} - {y^2} = {(a\sec \theta + b\tan \theta )^2} - {(a\tan \theta + b\sec \theta )^2} $ .
By applying the formula we will solve it now,
 $ \Rightarrow {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 $ $ = {a^2}{\sec ^2}\theta - {a^2}{\tan ^2}\theta + {b^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta \\ \Rightarrow {a^2}({\sec ^2}\theta - {\tan ^2}\theta ) + {b^2}({\sec ^2}\theta - {\tan ^2}\theta ) $
We will take the common factors together and we have:
 $ \Rightarrow ({\sec ^2}\theta - {\tan ^2}\theta )({a^2} - {b^2}) $ .
We know the trigonometric identity which says that $ {\sec ^2}\theta - {\tan ^2}\theta = 1 $ . So by substituting the value we get, $ 1({a^2} - {b^2}) $ .
Hence we get that $ {x^2} - {y^2} = {a^2} - {b^2} $ .
So, the correct answer is “ $ {x^2} - {y^2} = {a^2} - {b^2} $ ”.

Note: We should be careful while solving this kind of question. We should be aware of the trigonometric identities, their functions and rules and be careful while doing the calculation and also check the positive and the negative signs of the functions. We should also know the relation that $ {\sec ^2}\theta = 1 + {\tan ^2}\theta $ .