Question

show that ...
${(sec \: a - tan \: a)}^{2} = \frac{1 - sin \: a}{1 + sin \: a}$
​

Answer 1

Answer:

solve

Step-by-step explanation:

see attachments.

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: Show that (sec a - tan a)² = (1 - sin a)/(1 + sin a)

Solution: R.H.S. =  (1 - sin a)/(1 + sin a)

=  (1 - sin a)/(1 + sin a) ×  (1 - sin a)/(1 - sin a)

=  (1 - sin a)²/(1 - sin² a)    

= (1 - sin a)²/(cos² a)

= (1 - sin a)²/(cos a)²

= [(1/cos a) - (sin a/cos a)]²

= (sec a - tan a)²

= L.H.S.

Concept used:

1) sin² a + cos² a = 1

⇒ cos² a = 1 - sin² a

2) cos² a = (cos a)²

3) sec a = 1/cos a

4) tan a = sin a/cos a