Question

(1 + cot theta + tan theta )(sin theta - cos theta) is equal to secant theta upon cosec squared theta minus cosec squared theta upon secant squared theta


Prove it​

Answer 1

$\sf\huge\bold{Answer:}$

Given :

$⇒(1 + \cot \theta + \tan \theta)( \sin \theta - \cos \theta) = \frac{ \sec \theta }{ \cosec {}^{2} \theta } - \frac{ \cosec \theta}{ \sec {}^{2} \theta}$

To prove :

LHS=RHS

Solution :

$\fbox{LHS}$

$= (1 + \cot \theta + \tan \theta)( \sin \theta - \cos \theta)$

Using

$⇒ \cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \\ ⇒ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

we get,

$= (1 + \frac{ \cos \theta }{ \sin \theta } + \frac{ \sin \theta }{ \cos \theta } )( \sin \theta - \cos \theta)$

Now,on multiplying both terms

$= ( \sin \theta + \cos \theta + \frac{ { \sin }^{2} \theta }{ \cos \theta } - \cos \theta - \frac{ \cos {}^{2} \theta }{ \sin \theta} - \sin \theta)$

On simplifying ,we get

$= \frac{ { \sin }^{2} \theta }{ \cos \theta } - \frac{ \cos {}^{2} \theta }{ \sin \theta}$

Now,using

$⇒ \sin( \alpha ) = \frac{1}{ \cosec( \alpha ) } \\ ⇒ \cos( \alpha ) = \frac{1}{ \sec( \alpha ) } \\ ⇒ \sin {}^{2} ( \alpha ) = \frac{1}{ \cosec {}^{2} ( \alpha ) } \\ ⇒ \cos {}^{2} ( \alpha ) = \frac{1}{ \sec {}^{2} ( \alpha ) }$

We get,

$= \frac{ \sec \theta }{ \cosec {}^{2} \theta } - \frac{ \cosec \theta }{ \sec {}^{2}\theta }$

$\fbox{RHS}$

$= \frac{ \sec \theta }{ \cosec {}^{2} \theta } - \frac{ \cosec \theta }{ \sec {}^{2}\theta }$

$\boxed{ \boxed{LHS=RHS} }$

Hence proved !