**Evaluate the following**

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan^{2} 45° + cos^{2} 30° − sin^{2} 60°

(iii) ^{(cos45}°^{)}/_{(sec30}°_{+cosec30}°_{)}

(iv)

(v)

Solution:

(i) sin60° cos30° + sin30° cos 60°

= ^{ √3}/_{2}^{ }x ^{√3}/_{2} + ^{1}/_{2} x ^{½}

= ^{3}/_{4} + ^{1}/_{4}

= ^{4}/_{4}

= 1

(ii) 2tan^{2} 45° + cos^{2} 30° − sin^{2} 60°

= 2(1)^{2} + ^{3}/_{4} – ^{3}/_{4}

= 2

(iii) ^{(cos45}°^{)}/_{(sec30}°_{+cosec30}°_{)}

(iv)

(v)

**2. Choose the correct option and justify your choice. **

**(i) ^{(2tan30˚)}/_{(1+tan230˚)}**

**(a) sin 60˚**

**(b) cos 60˚**

**© tan60˚**

**(d) sin30˚**

**(ii) (1-tan ^{2}45˚)/(1+tan^{2}45˚)**

**(a) tan90˚**

**(b) 1˚**

**© sin45˚**

**(d) 0˚**

**(iii) sin2A = 2sinA is true when A =**

**(a) 0˚**

**(b) 30˚**

**© 45˚**

**(d) 60˚**

**(iv) 2tan30˚/(1-tan ^{2}30˚)**

**(a) cos60˚**

**(b) sin 60˚**

**©tan60˚**

**(d) sin30˚**

Solution:

**(i) ^{(2tan30˚)}/_{(1+tan²30˚)}**

Out of the given alternatives, only sin60˚= ^{√3}/_{2} .

Hence, (A) is correct.

**(ii) (1-tan ^{2}45˚)/(1+tan^{2}45˚)**

= [(1-(1)^{2}]/ [(1+(1)^{2}]

= (1-1)/(1+1)

= 0/2

= 0

Hence, (D) is correct.

(iii)**sin2A = 2sinA is true when A =**

Out of the given alternatives, only A = 0° is correct.

As sin 2A = sin 0° = 0

2sinA = 2sin 0° = 2(0) = 0. Hence, (A) is correct.

**(iv) 2tan30˚/(1-tan ^{2}30˚)**

Out of the given alternatives, only tan 60° = √3. Hence, (C) is correct.

**3. If tan(A+B) = √3 and tan(A-B) = ^{1}/_{√3}**

**0˚< A+B ≤ 90˚, A>B, find A and B.**

Solution:

tan(A+B) = √3

tan(A+B) = tan60

A+B = 60 ———(1)

tan(A-B) = ^{1}/_{√3}

⇒tan (A − B) = tan30

⇒A − B = 30 ————–(2)

On adding both equations, we obtain 2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

**4. State whether the following are true or false. Justify your answer. **

**(i) sin (A + B) = sin A + sin B **

**(ii) The value of sinθ increases as θ increases **

**(iii) The value of cos θ increases as θ increases **

**(iv) sinθ = cos θ for all values of θ **

**(v) cot A is not defined for A = 0° **

Solution:

(i) sin (A + B) = sin A + sin B

Let A = 30° and B = 60° ;

sin (A + B) = sin (30° + 60°) = sin 90° = 1

We have, sin A + sin B = sin 30° + sin 60°

^{1}/_{2} + ^{√3}/_{2} = ^{(1+√3)}/_{2}

Clearly, sin (A + B) ≠ sin A + sin B

Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as sin 0° = 0

sin 30˚= ^{1}/_{2} = 0.5

sin 45˚= ^{1}/_{√2} = 0.707

cos60˚= ^{1}/_{2} = 0.5

cos90˚ = 0

It can be observed that the value of cos θ does not increase in the interval of 0°<θ

<90°. Hence, the given statement is false. >

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45°

As sin˚45 = ^{1}/_{√2}

cos˚45 = ^{1}/_{√2}

It is not true for all other values of θ.

As sin30˚ = ^{1}/_{2 }and cos30 ˚ = ^{√3}/_{2}

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As CotA= cosA/sinA

cot0˚ = cos0˚/sin0˚ = ^{1}/_{0} undefined

Hence, the given statement is true.

We loved, put posts of logarithms

Yeah sure