Question

$\frac{tana}{seca - 1} + \frac{tana}{seca + 1} = 2coseca$
solve this fast​

Answer 1

Solution:-

$\rm \: \implies\dfrac{ \tan \theta }{ \sec \theta - 1} + \dfrac{ \tan\theta }{ \sec \theta+ 1 } = 2 \csc\theta$

$\rm \implies \: \dfrac{ \dfrac{ \sin\theta }{ \cos \theta} }{ \dfrac{1}{ \cos \theta} - 1 } + \dfrac{ \dfrac{ \sin\theta }{ \cos \theta} }{ \dfrac{1}{ \cos \theta} + 1 }$

$\rm \implies \dfrac{ \dfrac{ \sin \theta}{ \cos \theta} }{ \dfrac{1 - \cos \theta}{ \cos \theta} } + \dfrac{ \dfrac{ \sin \theta}{ \cos \theta} }{ \dfrac{1 + \cos \theta}{ \cos \theta} }$

$\rm \implies \dfrac{ \sin \theta }{ \cos \theta} \times \dfrac{ \cos \theta}{1 - \cos \theta} + \dfrac{ \sin \theta }{ \cos \theta} \times \dfrac{ \cos \theta}{1 + \cos \theta}$

$\rm \implies \dfrac{ \sin \theta }{ \cancel{\cos \theta}} \times \dfrac{ \cancel{ \cos \theta}}{1 - \cos \theta} + \dfrac{ \sin \theta }{ \cancel{\cos \theta}} \times \dfrac{ \cancel{\cos \theta}}{1 + \cos \theta}$

$\rm \implies \dfrac{ \sin \theta}{1 - \cos \theta} + \dfrac{ \sin \theta}{1 + \cos \theta}$

Taking Lcm

$\rm \implies \dfrac{ \sin \theta(1 + \cos \theta) + sin \theta(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$

$\rm \implies \: \dfrac{ \sin \theta + \sin \theta \cos \theta + \sin \theta - \sin \theta \cos \theta}{1 - \cos {}^{2} \theta}$

$\rm \implies \: \dfrac{ \sin \theta + \cancel{\sin \theta \cos \theta }+ \sin \theta - \cancel{\sin \theta \cos \theta}}{1 - \cos {}^{2} \theta}$

$\rm \implies \dfrac{2 \sin \theta}{ \sin {}^{2} \theta}$

$\rm \implies \dfrac{2}{ \sin \theta} = 2 \csc \theta$

Hence proved

Answer 2

First we simplify LHS

LHS

$\frac{tana}{seca - 1} + \frac{tana}{seca + 1}$

Take tanA common from the equation it is easier to do that question by taking tanA common

$tana( \frac{1}{seca - 1} + \frac{1}{seca + 1} ) \\ = tana( \frac{seca + 1 + seca - 1}{(seca - 1)(seca + 1)})$

Now in denominator use the identity (a+b)(a-b)=a^2-b^2

$=> tana( \frac{2seca}{ {sec}^{2} a - {1}^{2} }) = tana \times \frac{2seca}{ {tan}^{2}a }$

Here we use sec^2A-1=tan^2A now we use

secA=1/cosA and tanA=sinA/cosA

$\frac{2seca}{tana} = \frac{ \frac{2}{cosa} }{ \frac{sina}{cosa} } \\ => \frac{2}{cosa } \times \frac{cosa}{sina} = \frac{2}{sina}$

Now we use 1/sinA=2cosecA

$= 2coseca$

LHS=2cosecA

RHS=2cosecA

LHS=RHS

Hence proved