Question

plz give ans plzz ...it was very important to me​

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 Identities:

1. sin²θ + cos²θ = 1

2. sec²θ - tan²θ = 1

3. cosec²θ - cot²θ = 1

Given question: Prove that- 2 cos² θ + $\frac{2}{1 \: + \: cot^{2} \theta}$ = 2.

Solution: L.H.S = 2 cos² θ + $\frac{2}{1 \: + \: cot^{2} \theta}$

   = 2 cos² θ + $\frac{2}{cosec^{2} \theta}$

   = 2 cos² θ + 2 sin² θ

   = 2 (cos² θ + sin² θ)

   = 2 × 1

   = 2

   = R.H.S.

Formulae Used:

1) cosec²θ - cot²θ = 1

⇒ cosec²θ = 1 + cot²θ

2) cosec²θ = $\frac{1}{sin^{2} \theta}$

3)  sin²θ + cos²θ = 1

Answer 2

To ᴘʀoᴠᴇ:

• 2cos²θ + [ 2/( 1 + cot²θ ) ] = 2

Sᴏʟᴜᴛɪᴏɴ:

2cos²θ + [ 2/( 1 + cot²θ ) ] = 2

Taking LHS

And simplifying by using

1 + cot²θ = cosec²θ

→ 2cos²θ + [ 2/cosec²θ ]

→ 2cos²θ + [ 2 × ( 1/cosec²θ ) ]

We know that

1/cosecθ = sinθ

→ 2cos²θ + 2sin²θ

Taking common

→ 2( cos²θ + sin²θ )

Simplifying using first identity

sin²θ + cos²θ = 1

→ 2 ( 1 ) = 2

Comparing with RHS

2 = 2

LHS = RHS

Hence proved

Mᴏʀᴇ ɪɴꜰᴏʀᴍᴀᴛɪᴏɴ:

• sin²θ + cos²θ = 1

→ sin²θ = 1 - cos²θ

→ sinθ = ±√1 - cos²θ

→ cos²θ = 1 - sin²θ

→ cosθ = ±√1 - sin²θ

• sec²θ - tan²θ = 1

→ sec²θ = 1 + tan²θ

→ secθ = ±√1 + tan²θ

→ tan²θ = sec²θ - 1

→ tanθ = ±√sec²θ - 1

• cosec²θ - cot²θ = 1

→ cosec²θ = 1 + cot²θ

→ cosecθ = ±√1 + cot²θ

→ cot²θ = cosec²θ - 1

→ cotθ = ±√cosec²θ - 1