Question

Prove that sec^2x-sin^2x/tan^2x=cosec^2x-cos^2x

Answer 1

If x is any angle, solution :

$\text{To prove :} \dfrac{ \sec {}^{2} x - \sin {}^{2}x }{ \tan {}^{2} x} = \cosec {}^{2} x - { \cos}^{2} x$


Solving left hand side,

$\implies\dfrac{ \sec {}^{2} x - \sin {}^{2}x }{ \tan {}^{2} x} \\ \\ \\ \implies \dfrac{ \sec {}^{2} x }{ \tan {}^{2}x} - \dfrac{\sin {}^{2}x }{ \tan {}^{2} x}$


===========================
From the properties of trigonometry, we know : -
• sec²A = $\dfrac{1}{cos^2 A}$

• tan²A = $\dfrac{sin^2A}{cos^2 A}$
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$\implies \dfrac{ \dfrac{1}{ \cos {}^{2}x } }{ \dfrac{ \sin {}^{2}x }{ \cos {}^{2}x } } - \dfrac{\sin {}^{2}x }{ \dfrac{ \sin {}^{2} x}{ \cos {}^{2} x}} \\ \\ \\ \implies \bigg \{ \dfrac{1}{ \cos {}^{2} x} \times \dfrac{ { \cos}^{2} x}{ { \sin}^{2} x} \bigg \} - \bigg \{\dfrac{ { \sin}^{2}x }{1} \times \dfrac{ { \cos}^{2}x }{ { \sin }^{2} x} \bigg \} \\ \\ \\ \implies \frac{1}{ { \sin}^{2} x} - { \cos}^{2} x \\ \\ \\ \implies { \cosec}^{2} x - { \cos}^{2} x$



Hence,
$\dfrac{ \sec {}^{2} x - \sin {}^{2}x }{ \tan {}^{2} x} = \cosec {}^{2} x - { \cos}^{2} x$


Proved.

Answer 2

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