Question

Prve that

(tanθ+secθ-1)/(tanθ-secθ+1) = (1+sinθ)/(cosθ)​

Answer 1

Answer:

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Answer 2

To Prove :

• $\bf\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=\dfrac{1+sin\theta}{cos\theta}$

Solution :

In the given equation ;

• $\bf{LHS=\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}}$

• $\bf{RHS=\dfrac{1+sin\theta}{cos\theta}}$

Let us solve the LHS ,

$\\ \longrightarrow\bf \: \dfrac{ tan \theta + sec \theta - 1 }{tan \theta - sec \theta + 1}$

• We know that ;

• $\bf{sec^2\theta-tan^2\theta=1}$

• By replacing the value we get ,

$\\ \longrightarrow \bf \: \dfrac{ tan \theta + sec \theta - (sec {}^{2} \theta - tan {}^{2} \theta) }{tan \theta - sec \theta + 1}$

• using the identity a² - b² = (a+b)(a-b) we get ,

$\\ \longrightarrow\bf \:\dfrac{ tan \theta + sec \theta - (sec \theta + tan \theta)( sec\theta - tan\theta) }{tan \theta - sec \theta + 1}$

• By taking (tanθ +secθ) common in numertaor we get ,

$\\ \longrightarrow\bf \dfrac{( tan \theta + sec \theta)[ 1- (sec\theta - tan\theta)]}{tan \theta - sec \theta + 1} \\ \\ \\ \\ \longrightarrow \bf\dfrac{( tan \theta + sec \theta)[ ( - sec\theta + tan\theta) + 1]}{tan \theta - sec \theta + 1} \\ \\ \\ \\ \longrightarrow\bf \: \dfrac{( tan \theta + sec \theta)[ (tan\theta - sec\theta) + 1]}{tan \theta - sec \theta + 1}$

• Cancelling [(tanθ-secθ) +1] we get ;

$\\ \longrightarrow \: \bf \:( tan \theta + sec\theta)$

We know that ,

• $\bf{tan\theta=\dfrac{sin\theta}{cos\theta}}$

• $\bf{sec\theta=\dfrac{1}{cos\theta}}$

• By replacing the values we get ;

$\\ \longrightarrow \bf \: \frac{sin \theta}{cos \theta} + \frac{1}{cos \theta}$

• Since denominators are equal ;

$\\ \longrightarrow \bf \frac{sin \theta + 1}{cos\theta} \\ \\ \\ \\ \longrightarrow \bf \: \frac{1 + sin \theta}{cos \theta} \\ \\ \\ \\ \longrightarrow \large{ \sf{ RH S }}$

⠀⠀⠀⠀ HENCE PROVED