Question

please answer this asps​

Answer 1

Solution

In above attachment clearly solution is mentioned

Take LHS

$\dfrac{tan\:A\:+\: sec\:A\:-\:1}{tan\:A\:-\:sec\:A\:+\:1}$

$formula \implies \: tan^{2}\: +\: 1\: = \: sec^{2}\:A$

$sec^{2}\: A \: -\: tan^{2}\: A\: = \: 1$

$\implies \dfrac{tan\: A\: +\: sec\: A \: - \: (sec^{2}\:A\:-\:tan^{2}\:A}{tan\:A\: - \: sec\: A \: +\: 1}$

$\implies \dfrac{(tan\:A\: + \: sec\:A)\:-\:[(sec\:A\:-\:tan\:A)(sec\:A\: + \: tan\:A)]}{tan\:A\:-\: sec\: A\: +\: 1}$

$\implies \dfrac{(tan\:A\: + \: sec\:A)[1\: - \: (sec\: A\: - \tan\:A)]}{tan\:A\:-\: sec\: A\: +\: 1}$

$\implies \dfrac{(tan\:A\: + \: sec\:A)[1\: - \: sec\: A\: + \tan\:A]}{tan\:A\:-\: sec\: A\: +\: 1}$

Rearrange denominator for elimination

$\implies \dfrac{(tan\:A\: + \: sec\:A)[1\: - \: sec\: a\: + \tan\:A]}{1\:-\: sec\: A\: +\:tan\:A }$

$\implies tan\:A\: + \: sec\:A$

$\implies \dfrac{sin\: A}{cos\: A}\: + \: \dfrac{1}{cos\: A}$

$\implies \dfrac{sin\:A\: + \: 1}{cos\: A}$

or

$\implies \dfrac{1\: + \:sin\:A}{cos\: A}$

$\boxed{LHS = RHS}$

Hence Proved

Answer 2

Answer- The above question is from the chapter 'Introduction to Trigonometry'.

Trigonometry- The branch of Mathematics which helps in dealing with measure of three sides of a right-angled triangle is called Trigonometry.

Trigonometric Ratios:

sin θ  = Perpendicular/Hypotenuse

cos θ = Base/Hypotenuse

tan θ = Perpendicular/Base

cosec θ = Hypotenuse/Perpendicular

sec θ = Hypotenuse/Base

cot θ = Base/Perpendicular

Also, tan θ = sin θ/cos θ and cot θ = cos θ/sinθ.

Trigonometric Identites:

1. sin²θ + cos²θ = 1

2. sec²θ - tan²θ = 1

3. cosec²θ - cot²θ = 1

Given question: Prove that-

$\frac{tan \: A + sec \: A - 1}{tan \: A - sec \: A + 1}$ = $\frac{1 + sin \: A}{cos \: A}$

Solution: The answer has been attached.