Question

Question for Only:- ❏ Moderators ❏ Brainly Stars ❏ Best users ✰ 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

Given :-

• L.H.S = (1 + cotA + tanA)(sinA - CosA)
• R.H.S = SecA/Cosec²A - CosecA/sec²A

To Show :-

• Left hand side = Right hand side :-]

Solution :-

To show Left hand side is equal to Right hand side , at first we have to simplify L.H.S by applying trigonometry formula.

Calculation for L.H.S = R.H.S :-

⇒(1 + cotA + tanA)(sinA - CosA)

⇒(1 + CosA/sinA + sinA/cosA)(sinA - cosA)

⇒(sinA.cosA + cos²A + sin²/cosA.sinA)(sinA.cosA)

⇒(sinA.cosA + cos²A + sin²)×(sinA - cosA)/(sinA.cosA)

⇒sin³A - cos³A/sinA.cosA

⇒sin³A/sinA.cosA - cos³A/sinA.cosA

⇒ sin²A/cosA - cos²A/sinA

⇒ 1/cosA × sin²A - 1/sinA × cos²A

• 1/cosA = SecA. sinA = 1/cosecA

⇒ secA × 1/cosec²A - cosecA × 1/sec²A

⇒ SecA/cosec²A - cosecA/sec²A)

Hence, proved , L.H.S. = R.H.S

Answer 2

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

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 !