Question

can u please solve it and explain step by step. need to prove​

Answer 1

Question:-

$\rm \: prove \: \: \: \cot {}^{2} x - \frac{1}{ \sin {}^{2} x} + 1 = 0$

Solution:-

$\rm \: \: \: \: \cot {}^{2} x - \frac{1}{ \sin {}^{2} x} + 1 = 0$

We can write

$\rm \: \: \: \: \cot {}^{2} x + 1- \frac{1}{ \sin {}^{2} x} = 0$

Using this trigonometry identities

$\to \rm \: 1 + \cot {}^{2} ( \theta) = \csc {}^{2} ( \theta)$

$\to \rm \: \csc( \theta) = \frac{1}{ \sin( \theta) }$

Applying this identity, we get

$\rm \: \: \: \: \csc {}^{2} x - \frac{1}{ \sin {}^{2} x} = 0$

$\rm \: \: \: \: \frac{1}{ \sin {}^{2} x} - \frac{1}{ \sin {}^{2} x} = 0$

$\rm \: 0 = 0$

Hence proved

some Trigonometry identity

$\to\rm \: \sin {}^{2} (x) + \cos {}^{2} (x) = 1$

$\rm \: \to \: 1 + \cot {}^{2} (x) = \csc {}^{2} (x)$

$\rm \to \: 1 + \tan{}^{2} (x) = \sec {}^{2} (x)$

$\to \rm \: \tan(x) = \frac{ \sin(x) }{ \cos(x) }$

$\rm \: \to \: \cot(x) = \frac{ \cos(x) }{ \sin(x) }$

$\rm \: \to \csc(x) = \frac{1}{ \sin( x{}^{} ) }$

$\rm \: \to \sec(x) = \frac{1}{ \cos(x) }$