Question

tana-sina/sin^3A=secA/(1+cosa)​

Answer 1

AnswEr :

• To Prove :

$\dfrac{ \tan( \alpha ) - \sin( \alpha ) }{ \sin^{3} ( \alpha ) } = \dfrac{ \sec( \alpha ) }{1 + \cos( \alpha ) }$

• Proof :

I'm taking $(\alpha)$ at the Place of A.

$\longrightarrow \dfrac{ \tan( \alpha ) - \sin( \alpha ) }{ \sin^{3} ( \alpha) }$

$\longrightarrow ( \tan( \alpha ) - \sin( \alpha ) ) \times \dfrac{ 1}{ \sin^{3} ( \alpha) }$

⠀⠀⠀⠀⋆ tan(θ) = sin(θ) / cos(θ)

$\longrightarrow \bigg(\dfrac{ \sin( \alpha ) }{ \cos( \alpha ) } - \sin( \alpha ) \bigg) \times \dfrac{ 1}{ \sin^{3} ( \alpha) }$

$\longrightarrow \bigg(\dfrac{ 1}{ \cos( \alpha ) } -1 \bigg) \cancel{\sin( \alpha )} \times \dfrac{ 1}{ \cancel{\sin^{3} ( \alpha)} }$

$\longrightarrow \bigg(\dfrac{ 1}{ \cos( \alpha ) } -1 \bigg) \times \dfrac{ 1}{ \sin^{2} ( \alpha)}$

$\longrightarrow \bigg(\dfrac{ 1 - \cos( \alpha ) }{ \cos( \alpha ) } \bigg) \times \dfrac{ 1}{ \sin^{2} ( \alpha)}$

$\longrightarrow \dfrac{ 1 - \cos( \alpha ) }{ \cos( \alpha )\sin^{2} ( \alpha) }$

⠀⠀⠀⠀⋆ sin²(θ) = (1 - cos²(θ))

$\longrightarrow \dfrac{ 1 - \cos( \alpha ) }{ \cos( \alpha )(1 - \cos^{2}( \alpha )) }$

⠀⠀⠀⠀⋆ (a² - b²) = (a + b)(a - b)

$\longrightarrow \dfrac{ \cancel{1 - \cos( \alpha )} }{ \cos( \alpha )(1 + \cos( \alpha)) \cancel{(1 - \cos( \alpha ))} }$

$\longrightarrow \dfrac{1}{ \cos( \alpha )(1 + \cos( \alpha )) }$

$\longrightarrow \dfrac{1}{ \cos( \alpha ) } \times \dfrac{1}{(1 + \cos( \alpha )) }$

⠀⠀⠀⠀⋆ 1 / cos(θ) = sec(θ)

$\longrightarrow \sec( \alpha ) \times \dfrac{1}{(1 + \cos( \alpha )) }$

$\longrightarrow \large\dfrac{ \sec( \alpha ) }{(1 + \cos( \alpha )) }$

⠀

$\therefore \boxed{ \dfrac{ \tan( \alpha ) - \sin( \alpha ) }{ \sin^{3} ( \alpha ) } = \dfrac{ \sec( \alpha ) }{1 + \cos( \alpha ) } }$

Answer 2

Question :-

Prove that

$\implies \: \frac{ \tan a - \sin a}{ { \sin}^{3} a} = \frac{ \sec a}{1 + \cos a}$

Formula used :-

$\star \: \tan \theta = \frac{ \sin \theta}{ \cos \theta} \\ \star \sec \theta = \frac{1}{ \cos \theta}$

Proof :-

Firstly taking left hand side

$\implies \: \frac{ \tan a - \sin a}{ { \sin}^{3} a} \: \\ \\ \rightarrow \: \frac{ \frac{ \sin a}{ \cos a} - \sin a}{ { \sin}^{3} a} \\ \\ \rightarrow \: \frac{ \sin a - \sin a \: \cos a}{ \cos a \: { \sin}^{3} a} \\ \\ \rightarrow \frac{ \sin a(1 - \cos a)}{ \cos a \: { \sin}^{2} a} \\ \\ \rightarrow \: \frac{1 - \cos a}{ \cos a(1 - { \cos}^{2}a)} \: \: \\ \\ \because \sf{ {x}^{2} - {y}^{2} = (x + y)(x - y)} \\ \\ \rightarrow \: \frac{1 - \cos a}{ \cos a(1 - \cos a)(1 + \cos a)} \\ \\ \rightarrow \: \frac{1}{ \cos a} . \frac{1}{(1 + \cos a)} \\ \\ \rightarrow \: \frac{ \sec a}{1 + \cos a}$

left hand side = right hand side

hence proved .