Question

$\frac{cos \: a}{1 - tan \: a} + \frac{sin \: a}{1 - cot \: a} = sin \: a + cos \: a$
prove L.H.S = R.H.S​

Answer 1

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: cos a/(1 - tan a) + sin a/(1 - cot a) = sin a + cos a

Solution: L.H.S = cos a/(1 - tan a) + sin a/(1 - cot a)

= [cos a/(1 - tan a) × (1 + tan a)/(1 + tan a)] + [sin a/(1 - cot a) × (1 + cot a)/(1 + cot a)]

= [cos a(1 + tan a)/(1 - tan² a)] + [sin a(1 + cot a)/(1 - cot² a)]

= [cos a(1 + tan a)/sec² a] + [sin a(1 + cot a)/cosec² a]

= cos³ a(1 + tan a) + sin³ a(1 + cot a)

= cos³ a + (tan a × cos³ a) + sin³a + (cot a × sin³ a)

= cos³ a + sin³ a + (sin a . cos² a) + (cos a . sin² a)

= cos³ a + sin³ a + (sin a . 1 - sin² a) + (cos a . 1 - cos² a)

= cos³ a + sin³ a + sin a - sin³ a + cos a - cos³ a

= sin a + cos a

= R.H.S.

Concept used:

1) sec² a - tan² a = 1

⇒ sec² a = 1 + tan² a

2) cosec² a - cot² a = 1

⇒ cosec² a = 1 + cot² a

3) tan a = sin a/cos a

4) cot a = cos a/sin a

5) sin² a + cos² = 1

⇒ sin² a = 1 - cos² a

⇒ cos² a = 1 - sin² a