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Given:

sin(A + B ) = cos(A - B) = √3/2

Calculation:

sin(A + B) = √3/2

⇒ sin(A + B )= sin60°

⇒  A + B = 60°      ….(i)

And, cos(A - B) = √3/2

⇒ cos(A - B) = cos30°

⇒  A – B = 30°      ….(ii)

Adding (i) and (ii)

We get, 2A = 90°

⇒  A = 45°

Putting in (i)

We get, 

⇒ 45° + B = 60°

⇒  B = 15°

∴ A and B are 45°and 15°

Short trick

sin(A + B )= sin60 

cos(A - B) = cos30°

So, A + B = 60° and A – B = 30°

Takes oprtion he different is 30° and sum is 60° 

So option B and D are satisfy but in option D difference is negative

Then option B is right

A= 45° and B = 15° 

Given:

cot A = k

Formula:

cot A = Base/Perpendicular

sin A = Perpendicular/Hypotenuse 

(Hypotenuse)² = (Base)² + (Perpendicular)²

Calculation:

cot A = k/1

⇒ Base/Perpendicular = k/1

⇒ Base = k and Perpendicular = 1

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

⇒ (Hypotenuse)2 = k2 + 12

⇒ Hypotenuse = √(1 + K²)

Now, sin A = 1/√(1 + k²)

Given:

cos (x - y) = √3/2

sin (x + y) = 1/2

Formula:

cos30° = √3/2

sin30° = 1/2

Calculation:

cos(x - y) = √3/2

⇒ cos(x - y) = cos30°

(x - y) = 30°     ---- (1)

sin(x + y) = 1/2

⇒ sin(x + y) = sin30°

(x + y) = 30    ---- (2)

Add equation (1) and equation (2), we get

2x = 60°

∴ x = 30°

Given:

secθ = 20/16

Concept used:

sinθ = opposite/hypotenuse

Tanθ = opposite/adjacent 

Calculation:

⇒ secθ = 20/16

⇒ tanθ = 12/16,  sinθ = 12/20

⇒ tanθ = 3/4,  sinθ = 3/5

⇒ sinθ + tanθ 

⇒ (3/5) + (3/4)

⇒ 27/20

∴ The value of sinθ + tanθ is 27/20.

Given : 

xcosθ - ysinθ = 2 

xsinθ + ycosθ = 4 

Formula used: 

sin²θ + cos²θ = 1

(a + b)² = a² + b² + 2ab

Calculation: 

xcosθ - ysinθ = 2 

On squaring we get,

⇒ x²cos²θ + y²sin²θ – 2(xcosθ)(ysinθ) = 4      ….(i)

xsinθ + ycosθ = 4 

On squaring we get,

⇒ x²sin²θ + y²cos²θ + 2(ycosθ)(xsinθ) = 16      ….(ii)

On adding eqns (i) and (ii)

⇒ x²cos²θ + y²sin²θ – 2(xcosθ)(ysinθ) + x²sin²θ + y²cos²θ + 2(ycosθ)(xsinθ) = 4 + 16

⇒ x²(cos²θ + sin²θ) + y²(sin²θ + cos²θ) = 20

⇒ x² + y² = 20

∴ The correct answer is x² + y² = 20

Given:

A + B = 45° 

Concept used:

1.) cos(A + B) = cosAcosB – sinAsinB

2.) If A + B = 45° then (1 + tanA)(1 + tanB)

Calculations:

sinASinB + (1 + tanA)(1 + tanB) – cosAcosB

⇒ (1 + tanA)(1 + tanB) – cosAcosB + sinAsinB

⇒ (1 + tanA)(1 + tanB) – (cosAcosB – sinAsinB)

⇒ (1 + tanA)(1 + tanB) – cos(A + B)

Here A + B = 45° 

⇒ 2 – cos(45°)

⇒ 2 – 1/√2 

⇒ (2√2 – 1)/√2

∴ The correct answer is (2√2 – 1)/√2

Given: 

cot A + cosec A = 3

Formula used :

cosec² A - cot² A = 1

cosec A - cot A   = 1/ ( cosec A + cot A)

cot A + cosec A = 3     ......(eq1)

cosec A - cot A  =1/3   .......(eq2) 

By solving eq 1 and eq 2 By elimination

2 cosec A = 3 + 1/3

cosec A = 5/3 = H / P

 \({H^2} = {P^2} + {B^2}\)

\({5^2} = {3^2} + {B^2}\)

B² = 25 - 9

B2 = 16

base = 4

cosA = 4/5

Given:

tanθ = 3/4 

Concept used:

1.) tanθ = P/B

2.) sinθ = P/H 

3.) cosθ = B/H

4.) Pythagoras Theorem 

H² = P² + B² 

5.) sin(90° + θ) = cosθ 

6.) Cos(90° + θ) = -sinθ 

Where, 

P → Perpendicular 

B → Base

H → Hypotenus 

Calculations:

tanθ = 3/4 = P/B 

H² = P² + B² 

⇒ H² = (3)² + (4)²

⇒ H = √25 = 5 

sinθ = 3/5 

cosθ = 4/5 

sin2(90° + θ) + tan2θ – cos2(90° + θ)

⇒ cos²θ + tan²θ – (-sinθ)²

⇒ cos2θ + tan2θ – sin²θ 

⇒ (4/5)² + (3/4)² – (3/5)² 

⇒ 16/25 + 9/16 – 9/25 

⇒ (256 + 225 – 144)/400

⇒ 337/400

∴ The correct answer is 337/400 

Concept:

cos A - cos B = 2 sin \(\rm (\frac {B + A}{2})\) sin \(\rm (\frac {B - A}{2})\) 

Calculations:

To express (cos 5x - cos 7x) as a product of sines or cosines or sines and cosines

we know the trigonometric formula,

cos A - cos B = 2 sin \(\rm (\frac {B + A}{2})\) sin \(\rm (\frac {B - A}{2})\) 

Here, A = 5x and B = 7x 

(cos 5x - cos 7x) = 2 sin \(\rm (\frac {7x + 5x}{2})\) sin \(\rm (\frac {7x - 5x}{2})\) 

(cos 5x - cos 7x) = 2 sin 6x sin x

Given:

Y = tan35°

Formula:

tan (90 - a) = cot a

Calculation:

⇒ cot 55° = cot(90 - 35) = tan 35°

⇒ tan 55° = tan(90 - 35) = cot 35 = 1/tan 35 = 1/Y

Then,

⇒ (2 tan55° + cot55°) = 2/Y + Y = (2 + Y²)/Y

∴ (2tan55° + cot55°) = (2 + Y²)/Y