Question

prove it with explanations and stepwise please answer my question I will mark them as brainlist ​

Answer 1

Question :

Prove that ;

$\sf{ \dfrac{ \tan \theta }{ \sec \theta - 1} + \dfrac{ \tan \theta}{ \sec \theta + 1} = 2 \csc \theta }$

Answer

Given : -

$\sf{ \dfrac{ \tan \theta }{ \sec \theta - 1} + \dfrac{ \tan \theta}{ \sec \theta + 1} = 2 \csc \theta }$

Required to prove : -

• LHS = RHS

Identity used : -

sec² θ - tan² θ = 1

Proof : -

$\sf{ \dfrac{ \tan \theta }{ \sec \theta - 1} + \dfrac{ \tan \theta}{ \sec \theta + 1} = 2 \csc \theta }$

We need to prove that LHS = RHS

So,

Consider the LHS part

$\sf{ \dfrac{ \tan \theta }{ \sec \theta - 1} + \dfrac{ \tan \theta}{ \sec \theta + 1} }$

Multiply the 1st one's numerator and denominator with sec θ + 1

$\sf{ \dfrac{ \tan \theta }{ \sec \theta - 1} \times \dfrac{ \sec \theta + 1}{ \sec \theta + 1} + \dfrac{ \tan \theta}{ \sec \theta + 1}} \\ \\ \\ \sf{ \dfrac{ \tan \theta( \sec \theta + 1)}{( \sec \theta - 1)( \sec \theta + 1)} + \dfrac{ \tan \theta}{ \sec \theta + 1} } \\ \\ \textsf{ Using the identity ; }\\ \mathtt{ ( a + b ) ( a - b ) = a^2 - b^2} \\ \\ \\ \sf{ \dfrac{ \tan \theta \sec \theta + \tan \theta}{ {\sec }^{2} \theta - 1 } + \dfrac{ \tan \theta}{ \sec \theta + 1} }$

Now,

Multiply the 2nd one's numerator and denominator with sec θ - 1

$\tt{ \dfrac{ \tan \theta \sec \theta + \tan \theta}{ { \sec}^{2} \theta - 1} + \dfrac{ \tan \theta}{ \sec \theta + 1} \times \dfrac{ \sec \theta - 1}{ \sec \theta - 1}} \\ \\ \\ \tt{ \dfrac{ \tan \theta \sec \theta + \tan \theta}{ { \sec}^{2} \theta - 1 } + \dfrac{ \tan \theta( \sec \theta + 1)}{( \sec \theta + 1)( \sec \theta - 1)} } \\ \\ \\ \dfrac{ \tan \theta \sec \theta + \tan \theta}{ { \sec}^{2} \theta - 1 } + \dfrac{ \tan \theta \sec \theta + \tan \theta}{ { \sec}^{2} - 1 }$

Using the identity ;

sec² θ - tan² θ = 1

sec² θ - 1 = tan² θ

$\dfrac{ \tan \theta \sec \theta + \tan \theta}{ { \tan}^{2} \theta} + \dfrac{ \tan \theta \sec \theta - \tan \theta}{ { \tan}^{2} \theta }$

Since,

• tan θ = sin θ/cos θ

• sec θ = 1/cos θ

$\dfrac{ \dfrac{ \sin \theta}{ \cos \theta} \times \dfrac{1}{ \cos \theta } + \dfrac{ \sin \theta}{ \cos \theta} }{ { \tan}^{2} \theta} + \dfrac{ \dfrac{ \sin \theta}{ \cos \theta} \times \dfrac{1}{ \cos \theta } - \dfrac{ \sin \theta}{ \cos \theta} }{ { \tan}^{2} \theta} \\ \\ \\ \dfrac{ \dfrac{ \sin \theta}{ { \cos}^{2} \theta } + \dfrac{ \sin \theta}{ \cos \theta } }{ { \tan}^{2} \theta } + \dfrac{ \dfrac{ \sin \theta}{ { \cos}^{2} \theta } - \dfrac{ \sin \theta}{ \cos \theta } }{ { \tan}^{2} \theta } \\ \\ \\ \dfrac{ \dfrac{ \sin \theta + \sin \theta \cos \theta}{ { \cos}^{2} \theta} }{ { \tan}^{2} \theta } + \dfrac{ \dfrac{ \sin \theta - \sin \theta \cos \theta}{ { \cos}^{2} \theta} }{ { \tan}^{2} \theta } \\ \\ \\ \\ \dfrac{ \dfrac{ \sin \theta + \sin \theta \cos \theta}{ { \cos}^{2} \theta} }{ \dfrac{ { \sin}^{2} \theta}{ { \cos }^{2} \theta} } + \dfrac{ \dfrac{ \sin \theta - \sin \theta \cos \theta}{ { \cos}^{2} \theta} }{ \dfrac{ { \sin}^{2} \theta}{ { \cos }^{2} \theta} }$

Now,

cos² θ get's cancelled since it is present in denominations of both fractions

$\dfrac{ \sin \theta + \sin \theta \cos \theta}{ { \sin}^{2} \theta } + \dfrac{ \sin \theta - \sin \theta \cos \theta}{ { \sin}^{2} \theta} \\ \\ \\ \sf \dfrac{ \sin \theta + \sin \theta \cos \theta + \sin \theta - \sin \theta \cos \theta}{ { \sin}^{2} \theta } \\ \\ \\ \sf \dfrac{ \sin \theta + \sin \theta}{ { \sin}^{2} \theta } \\ \\ \\ \sf \dfrac{2 \sin \theta}{ { \sin}^{2} \theta} \\ \\ \\ \tt 2 \times \dfrac{ \sin \theta}{ { \sin}^{2} \theta } \\ \\ \\ \tt 2 \times \dfrac{1}{ \sin \theta} \\ \\ \\ 2 \times \csc \theta \\ \\ \\ \tt \implies 2 \csc \theta$

Consider the RHS part

2 cosec θ

LHS = RHS

Hence Proved ✓

Answer 2

Answer :

$Generally \;to \;prove \;something\; in\; mathematics ,$

$We \;consider \;L.H.S\;( or) \:R.H.S \;and \;derive \;R.H.S \;and\; L.H.S\; respectively$

NOTE :

• $Here \;I\; represented\; \theta \;as\; x$

Considering L.H.S

• $\dfrac{tanx}{secx - 1} + \dfrac{tanx}{secx+1}$

• $Lets \;multiply \;and \;divide\;the\;first \;term \;with (secx+1)\; and\; second \;with \;(secx-1)$

=> $\dfrac{(tanx)}{(secx - 1)} \times\dfrac{(secx + 1)}{(secx+1)}$

+ $\dfrac{(tanx)}{(secx + 1)}\times \dfrac{(secx-1)}{(secx-1)}$

Reminder :

• $(a+b)(a-b)=a^2-b^2$

• $tan^2x = sec^2x - 1$

• $secx= \dfrac{1}{cosx}$

• $tanx =\dfrac{sinx}{cosx}$

• $cosecx =\dfrac{1}{sinx}$

Coming back to solution,

=> $\dfrac{(tanx)}{(secx - 1)} \times\dfrac{(secx + 1)}{(secx+1)}$

+ $\dfrac{(tanx)}{(secx + 1)}\times \dfrac{(secx-1)}{(secx-1)}$

=> $\dfrac{tanx(secx-1)}{sec²x - 1} + \dfrac{tanx(secx+1)}{sec²x - 1}$

=> $\dfrac{tanxsecx - tanx + tanxsecx + tanx}{sec²x-1}$

=> $\dfrac{2tanxsecx}{tan²x}$

=> $\dfrac{2secx}{tanx}$

=> $\dfrac{2(\dfrac{1}{(cosx)})}{(\dfrac{sinx}{cosx})}$

=> $\dfrac{2}{sinx}$

=> $2cosecx$

Therefore we derived R.H.S from L.H.S

Hence ,

$\dfrac{tanx}{secx - 1} + \dfrac{tanx}{secx + 1}$ = $2cosecx$

$Replacing\;x\; with\;\theta$

$\dfrac{tan\theta}{sec\theta- 1} + \dfrac{tan\theta}{sec\theta+ 1}$ = $2cosec\theta$

Proved !