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# If $a\sec \theta + b\tan \theta + c = 0$ and ${\text{p}}\sec \theta + q\tan \theta + r = 0$, prove that${\left( {br - qc} \right)^2} - {\left( {pc - ar} \right)^2} = {\left( {aq - bp} \right)^2} \\$  Hint: Make use of the cross multiplication technique & trigonometric identities to solve this problem.

$a \sec \theta + b\tan \theta + c = 0.........................(1) \\$

$p \sec \theta + q\tan \theta + r = 0..........................(2) \\$

Solve these two equations for $\sec \theta$ and $\tan \theta$ by the cross multiplication technique, we get,

$\dfrac{{\sec \theta }}{{br - qc}} = \dfrac{{\tan \theta }}{{cp - ar}} = \dfrac{1}{{aq - bp}} \\$

$\Rightarrow \sec \theta = \dfrac{{br - qc}}{{aq - bp}}$ & $\tan \theta = \dfrac{{cp - ar}}{{aq - bp}} \\$

Now, you know that,

${\sec^2}\theta - {\tan^2}\theta = 1 \\$

$\Rightarrow {\left( {\dfrac{{br - qc}}{{aq - bp}}} \right)^2} - {\left( {\dfrac{{cp - ar}}{{aq - bp}}} \right)^2} = 1 \\$

$\Rightarrow {\left( {br - qc} \right)^2} - {\left( {cp - ar} \right)^2} = {\left( {aq - bp} \right)^2} \\$

Hence Proved.

Note: In this particular type of question, solve these equations and then apply trigonometry identity to get your answer. Mistakes should be avoided during cross multiplication.

View Notes
Tan Theta Formula  Sin Theta Formula  Cos Theta Formula  Cos Square Theta Formula  Tangent 3 Theta Formula  Trigonometric Identities - Class 10  Cos 0  Value of Sin 0  Sec 0  Tan 0 Degrees  