Question

prove this trignometric identity please​

Answer 1

Answer:

(sec theta + tan theta -1)/(tan theta - sec theta +1)

= cos theta / 1-sin theta

LHS = (sec theta + tan theta -(sec²theta - tan²theta)/(tan theta - sec theta +1)

= (sec theta + tan theta)(1 - sec theta + tan theta)

/( tan theta - sec theta +1)

= sec theta + tan theta

= 1/ cos theta + sin theta /cos theta

= (1+sin theta)/cos theta

= (1+sin theta) x (1 -sin theta) / (cos theta x (1 -sin theta))

= (1-sin²theta)/ (cos theta x (1-sin theta))

= cos² theta/( cos theta x (1-sin theta))

= cos theta/ (1-sin theta)

hence proved....

Answer 2

Solution :-

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

Consider LHS

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

Substituting 1 = sec² θ - tan² θ in numerator

$\sf = \dfrac{sec \theta + tan \theta - (sec^{2} \theta - tan^{2} \theta) }{tan \theta - sec \theta + 1}$

Expanding sec² θ - tan² θ = (sec θ + tan θ)(sec θ - tan θ)

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

Taking (sec θ + tan θ) common in numerator

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

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

By rearranging the terms

$\sf = \dfrac{sec \theta + tan \theta( \cancel{1 - sec\theta + tan \theta}) }{ \cancel{1- sec \theta + tan \theta}}$

$\sf =sec \theta + tan \theta$

$\sf = \dfrac{1}{cos \theta} + \dfrac{sin \theta}{cos \theta}$

[ Because tanθ = sinθ/cosθ and secθ = 1/cosθ ]

$\sf = \dfrac{1 + sin \theta}{cos \theta}$

Multiplying by (1 - sinθ) with numerator and denominator

$\sf = \dfrac{(1 + sin \theta)(1 - sin \theta)}{cos \theta(1 - sin \theta)}$

$\sf = \dfrac{ {1}^{2} - sin^{2} \theta }{cos \theta(1 - sin \theta)}$

[ Because (a + b)(a - b) = a² - b² ]

$\sf = \dfrac{1- sin^{2} \theta }{cos \theta(1 - sin \theta)}$

$\sf = \dfrac{cos^{2} \theta }{cos \theta(1 - sin \theta)}$

[ Because 1 - sin²θ = cos²θ ]

$\sf = \dfrac{cos\theta }{1 - sin \theta}$

= RHS

Hence proved.