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Question 1:

Evaluate the following:

(i) $sin60^{\circ}cos30^{\circ}+sin30^{\circ}cos60^{\circ}$

(ii) $2tan^{2}45^{\circ}+cos^{2}30^{\circ}-sin^{2}60^{\circ}$

(iii) $\frac{cos45^{\circ}}{sec30^{\circ}+cosec60^{\circ}}$

(iv) $\frac{sin30^{\circ}+tan45^{\circ}-cosec60^{\circ}}{sec30^{\circ}+cos60^{\circ}+cot45^{\circ}}$

(v) $\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45^{\circ}}{sin^{2}30+cos^{2}30^{\circ}}$

Solution 1:

(i) sin60° cos30° + sin30° cos 60°

= $\left ( \frac{1}{2} \right )\left ( \frac{1}{2} \right )+\left ( \frac{\sqrt{3}}{2} \right )\left ( \frac{\sqrt{3}}{2} \right )$

$=\frac{1}{4}+\frac{3}{4}=1$

(ii) 2tan²45° + cos²30° − sin²60°

= $2(1)^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}-\left ( \frac{\sqrt{3}}{2} \right )^{2}$

= $2+\frac{3}{4}-\frac{3}{4}=2$

(iii) $\frac{cos45^{\circ}}{sec30^{\circ}+cosec60^{\circ}}$

$\frac{\frac{1}{\sqrt{2}}}{ \frac{2}{\sqrt{3}}+2}$ $=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}} }$

$=\frac{\sqrt{3}}{\sqrt{2}(2+2\sqrt{3})}=\frac{\sqrt{3}}{\sqrt{2}(2+2\sqrt{6})}$

= $\frac{\sqrt{3}(2\sqrt{6}-2\sqrt{2})}{\left ( 2\sqrt{6}+2\sqrt{2} \right )\left ( 2\sqrt{6}-2\sqrt{2} \right )}$

= $\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{ \left ( 2\sqrt{6} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}$ $==\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{ 24-8}$

$=\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{ 16}$ $=\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3\sqrt{2}-\sqrt{6}}{8}$

(iv) $\frac{sin30^{\circ}+tan45^{\circ}-cosec60^{\circ}}{sec30^{\circ}+cos60^{\circ}+cot45^{\circ}}$

$\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{ \frac{2}{\sqrt{3}}+\frac{1}{2}+1}$ $=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$

$\frac{\frac{3\sqrt{3}-4}{2\sqrt{3}}}{\frac{3\sqrt{3}+4}{2\sqrt{3}}}=\frac{(3\sqrt{3}-4)}{(3\sqrt{3}+4)}$

$=\frac{\left ( 3\sqrt{3}-4 \right )\left ( 3\sqrt{3}-4 \right )}{\left ( 3\sqrt{3}-4 \right )\left ( 3\sqrt{3}+4 \right )}$ $=\frac{\left ( 3\sqrt{3}-4 \right )^{2}}{\left ( 3\sqrt{3} \right )^{2}-(4)^{2}}$

$=\frac{43-24\sqrt{3}}{11}$

(v) $\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45^{\circ}}{sin^{2}30+cos^{2}30^{\circ}}$

$\frac{5\left ( \frac{1}{2} \right )^{2}+4\left ( \frac{2}{\sqrt{3}} \right )^{2}-(1)^{2}}{ \left ( \frac{1}{2} \right )^{2}+\left ( \frac{\sqrt{}3}{2} \right )^{2}}$

$=\frac{5\left ( \frac{1}{4} \right )+\left ( \frac{16}{3} \right )-1 }{\left ( \frac{1}{2} \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}}$ $=\frac{5\left ( \frac{1}{4} \right )+\left ( \frac{16}{3} \right )-1}{\frac{1}{4}+\frac{3}{4}}$

$=\frac{67}{12}$

Comments

Tasneem

Question 2:

Choose the correct option and justify your choice:

(i) $\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}$ =

(A) $sin60^{\circ}$ (B) $cos60^{\circ}$ (C) $tan60^{\circ}$ (D) $sin30^{\circ}$

(ii) $\frac{1-tan^{2}45^{\circ}}{1+tan^{2}45^{\circ}}=$

(A) $tan90^{\circ}$ (B) 1 (C) $sin45^{\circ}$ (D) 0

(iii) sin 2A = 2 sin A is true when A =

(A) $0^{\circ}$ (B) $30^{\circ}$ (C) $45^{\circ}$ (D) $60^{\circ}$

(iv) $\frac{2tan30^{\circ}}{1-tan^{2}30^{\circ}}=$

(A) $cos60^{\circ}$ (B) $sin60^{^{\circ}}$ (C) $tan60^{\circ}$ (D) $sin30^{\circ}$

Solution 2:

(i)

Out of the given alternatives, only

Hence, (A) is correct.

(ii)

Hence, (D) is correct.

(iii) As sin 2A = sin 0° = 0

2 sinA = 2sin 0° = 2(0) = 0

Hence, (A) is correct.

(iv)

Out of the given options, only tan 60°

Hence, (C) is correct.

Comments

Tasneem

Question 3:

If $tan(A+B)=\sqrt{3}$ and $tan(A-B)=\frac{1}{^{\sqrt{3}}}$ ; $0^{\circ}< A+B\leq 90^{\circ}$ ; A > B, find A and B.

Solution 3:

⇒

⇒ A + B = 60 … (1)

⇒ 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°

Comments

Tasneem

Question 4:

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

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

(ii) The value of $sin\Theta$ increases as $\Theta$ increases.

(iii) The value of $cos\Theta$ increases as $\Theta$ increases.

(iv) $sin\Theta =cos\Theta$ for all values of $\Theta$ .

(v) cot A is not defined for A = $0^{\circ}$ .

Solution 4:

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

Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°)

= sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

$=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}$

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

$sin30^{\circ}=\frac{1}{2}=0.5$

$sin45^{\circ}=\frac{1}{\sqrt{2}}=0.707$

$sin60^{\circ}=\frac{\sqrt{3}}{2}=0.866$

sin 90° = 1

Hence, the given statement is true.

(iii) cos 0° = 1

$cos30^{\circ}=\frac{\sqrt{3}}{2}=0.866$

$cos45^{\circ}=\frac{1}{\sqrt{2}}=0.707$

$cos60^{\circ}=\frac{1}{2}=0.5$

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°

It is not true for all other values of θ.

As and ,

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As ,

= undefined

Hence, the given statement is true.

Comments

Tasneem

All Exercises - Chapter 8 - Introduction to Trigonometry

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

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Learn Exercise 8.2 with Free Lessons & Tips

All Exercises - Chapter 8 - Introduction to Trigonometry

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