Question

If cosec A = $\frac{13}{5}$ then prove that ${\tan }^{2}A-{\sin }^{2}A={\sin }^{4}A{\sec }^{2}A$

Answer 1

$cosecA=\frac{13}{5}$

$\Rightarrow \sin A=\frac{5}{13}$

$\Rightarrow \cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-\frac{25}{169}}$

$=\sqrt{\frac{144}{169}}$

$=\frac{12}{13}$

$\Rightarrow \cos A=\frac{12}{13}$

$\Rightarrow \sec A=\frac{13}{12}$

$\Rightarrow \tan A=\frac{\sin A}{\cos A}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$

Now, ${\tan }^{2}A-{\sin }^{2}A={\left(\frac{5}{12}\right)}^2-{\left(\frac{5}{13}\right)}^2$

$=\frac{25}{144}-\frac{25}{169}$

$=25\left[\frac{169-144}{169\times 144}\right]$

$=\frac{25\times 25}{169\times 144}$

$\&$ ${\sin }^{2}A{\sec }^{2}A={\left(\frac{5}{13}\right)}^4\times \frac{{\left(13\right)}^2}{{\left(12\right)}^2}$

$=\frac{5\times 5\times 5\times 5}{{\left(13\right)}^2.{\left(12\right)}^2}$

$=\frac{25\times 25}{169\times 144}$

Hence, ${\tan }^{2}A-{\sin }^{2}A={\sin }^{4}A{\sec }^{4}A$

Hence, the answer is proved.