If 13sectheta - 5tantheta =13 then the value of 13tantheta - 5sectheta can be?

Mar 12, 2017

$\pm 5$

Explanation:

$\left\{\begin{matrix}\frac{a}{\cos} \theta - b \sin \frac{\theta}{\cos} \theta = a \\ - \frac{b}{\cos} \theta + a \sin \frac{\theta}{\cos} \theta = x\end{matrix}\right.$

after multiplying the first equation by $b$ and the second by $a$ with posterior add term to term we have

$\left({a}^{2} - {b}^{2}\right) \sin \frac{\theta}{\cos} \theta = a b + a x$

after multiplying the first equation by $a$ and the second by $b$ with posterior add term to term we have

$\frac{{a}^{2} - {b}^{2}}{\cos} \theta = {a}^{2} + b x$ then

$a b + a x = \left({a}^{2} + b x\right) \sin \theta$ so

$\left\{\begin{matrix}\sin \theta = \frac{a \left(b + x\right)}{{a}^{2} + b x} \\ \cos \theta = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + b x}\end{matrix}\right.$

but ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

then

${a}^{2} {\left(b + x\right)}^{2} + {\left({a}^{2} - {b}^{2}\right)}^{2} = {\left({a}^{2} + b x\right)}^{2}$

solving for $x$ we obtain

$x = \pm b = \pm 5$

Mar 12, 2017

Given

$13 \sec \theta - 5 \tan \theta = 13$

$\implies {\left(13 \sec \theta - 5 \tan \theta\right)}^{2} = {13}^{2}$

$\implies {13}^{2} {\sec}^{2} \theta + {5}^{2} {\tan}^{2} \theta - 2 \cdot 13 \cdot 5 \sec \theta \tan \theta = {13}^{2}$

$\implies {13}^{2} \left({\sec}^{2} \theta - 1\right) + {5}^{2} {\sec}^{2} \theta - {5}^{2} - 2 \cdot 13 \cdot 5 \sec \theta \tan \theta = 0$

$\implies {13}^{2} {\tan}^{2} \theta + {5}^{2} {\sec}^{2} \theta - 2 \cdot \left(13 \tan \theta\right) \cdot \left(5 \sec \theta\right) = {5}^{2}$

$\implies {\left(13 \tan \theta - 5 \sec \theta\right)}^{2} = {5}^{2}$

$\implies \left(13 \tan \theta - 5 \sec \theta\right) = \pm 5$