Chapter 8: Introduction to Trigonometry (Trigonometry)

$AC=\sqrt{24^2+7^2}=\sqrt{625}=25$ cm.

1. $\sin A=\frac{BC}{AC}=\frac{7}{25}$, $\cos A=\frac{AB}{AC}=\frac{24}{25}$
2. $\sin C=\frac{AB}{AC}=\frac{24}{25}$, $\cos C=\frac{BC}{AC}=\frac{7}{25}$

(i) sin A = 7/25, cos A = 24/25 (ii) sin C = 24/25, cos C = 7/25

$QR=\sqrt{13^2-12^2}=5$ cm. Then $\tan P=\frac{QR}{PQ}=\frac{5}{12}$ and $\cot R=\frac{QR}{PQ}=\frac{5}{12}$.

tan P – cot R = 0

Take opposite $=3k$, hypotenuse $=4k$; then adjacent $=\sqrt{16k^2-9k^2}=\sqrt{7}k$.

cos A = √7/4, tan A = 3/√7

$\cot A=\frac{8}{15}$: adjacent $8k$, opposite $15k$, hypotenuse $\sqrt{64+225}k=17k$.

sin A = 15/17, sec A = 17/8

Hypotenuse $13k$, adjacent $12k$, so opposite $=5k$.

sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5

Consider a right triangle ABC with the right angle at C. Then $\cos A=\frac{AC}{AB}$ and $\cos B=\frac{BC}{AB}$.

Since $\cos A=\cos B$, $AC=BC$. Angles opposite equal sides are equal, so $\angle B=\angle A$.

cos A = cos B gives AC = BC, so the angles opposite them are equal: ∠A = ∠B.

1. $\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta }=\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta =\frac{49}{64}$
2. ${\cot }^2\theta =\frac{49}{64}$

(i) 49/64 (ii) 49/64

$\tan A=\frac{3}{4}$, so the sides are 4, 3, 5: $\sin A=\frac{3}{5}$, $\cos A=\frac{4}{5}$.

• LHS $=\frac{1-9/16}{1+9/16}=\frac{7}{25}$
• RHS $=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

Yes, both sides equal 7/25.

$\tan A=\frac{1}{\sqrt{3}}$ gives $A=30°$, so $C=60°$.

1. $\frac{1}{2}\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1$
2. $\frac{\sqrt{3}}{2}\cdot \frac{1}{2}-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}=0$

(i) 1 (ii) 0

Let $QR=x$; then $PR=25-x$. By Pythagoras: $(25-x{)}^2=x^2+25\Rightarrow 625-50x=25\Rightarrow x=12$. So $QR=12$, $PR=13$.

sin P = 12/13, cos P = 5/13, tan P = 12/5

1. False. For example, $\tan 60°=\sqrt{3}>1$.
2. True. $\sec A=\frac{\text{hypotenuse}}{\text{adjacent}}$ and the hypotenuse is the longest side, so $\sec A\ge 1$; $\frac{12}{5}>1$ is possible (sides 5, 12, hypotenuse 12 and adjacent 5).
3. False. cos A means the cosine of A; cosecant is written cosec A.
4. False. "cot" alone has no meaning; cot A is one symbol, the cotangent of A.
5. False. $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\le 1$, but $\frac{4}{3}>1$.

(i) False (ii) True (iii) False (iv) False (v) False

1. $\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}+\frac{1}{4}=1$
2. $2(1)+\frac{3}{4}-\frac{3}{4}=2$
3. $\frac{1/\sqrt{2}}{\frac{2}{\sqrt{3}}+2}=\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})}=\frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}\times 2}=\frac{3-\sqrt{3}}{4\sqrt{2}}=\frac{3\sqrt{2}-\sqrt{6}}{8}$
4. $\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}$ (multiplying through by $2\sqrt{3}$) $=\frac{(3\sqrt{3}-4{)}^2}{27-16}=\frac{43-24\sqrt{3}}{11}$
5. $\frac{5\cdot \frac{1}{4}+4\cdot \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{5}{4}+\frac{16}{3}-1=\frac{67}{12}$

(i) 1 (ii) 2 (iii) (3√2 – √6)/8 (iv) (43 – 24√3)/11 (v) 67/12

1. $\frac{2/\sqrt{3}}{4/3}=\frac{\sqrt{3}}{2}=\sin 60°$: (A)
2. $\frac{1-1}{1+1}=0$: (D)
3. At $A=0°$: $\sin 0°=0=2\sin 0°$ ✓. At 30°: $\sin 60°=\frac{\sqrt{3}}{2}\ne 1$. (A)
4. $\frac{2/\sqrt{3}}{2/3}=\sqrt{3}=\tan 60°$: (C)

(i) (A) (ii) (D) (iii) (A) (iv) (C)

$A+B=60°$ and $A-B=30°$. Adding: $2A=90°$.

A = 45°, B = 15°

1. False. With $A=B=45°$: $\sin 90°=1$ but $\sin 45°+\sin 45°=\sqrt{2}$.
2. True (for $0°\le \theta \le 90°$): 0, 1/2, 1/√2, √3/2, 1.
3. False. cos θ decreases: 1, √3/2, 1/√2, 1/2, 0.
4. False. It holds only at $\theta =45°$.
5. True. $\cot 0°=\frac{\cos 0°}{\sin 0°}=\frac{1}{0}$, which is not defined.

(i) False (ii) True (iii) False (iv) False (v) True

From ${\mathrm{cosec}}^{2}A=1+{\cot }^{2}A$:

$$\sin A=\frac{1}{\sqrt{1+{\cot }^{2}A}},\tan A=\frac{1}{\cot A},\sec A=\sqrt{1+{\tan }^{2}A}=\frac{\sqrt{1+{\cot }^{2}A}}{\cot A}$$

sin A = 1/√(1 + cot²A), sec A = √(1 + cot²A)/cot A, tan A = 1/cot A

$$\cos A=\frac{1}{\sec A},\tan A=\sqrt{{\sec }^{2}A-1},\sin A=\tan A\cos A=\frac{\sqrt{{\sec }^{2}A-1}}{\sec A}$$

$$\cot A=\frac{1}{\sqrt{{\sec }^{2}A-1}},\mathrm{cosec}A=\frac{\sec A}{\sqrt{{\sec }^{2}A-1}}$$

cos A = 1/sec A, sin A = √(sec²A – 1)/sec A, tan A = √(sec²A – 1), cot A = 1/√(sec²A – 1), cosec A = sec A/√(sec²A – 1)

1. $9({\sec }^{2}A-{\tan }^{2}A)=9\times 1=9$: (B)
2. $\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)=\frac{(\sin \theta +\cos \theta {)}^2-1}{\sin \theta \cos \theta }=\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }=2$: (C)
3. $\frac{1+\sin A}{\cos A}(1-\sin A)=\frac{1-{\sin }^{2}A}{\cos A}=\frac{{\cos }^{2}A}{\cos A}=\cos A$: (D)
4. $\frac{{\sec }^{2}A}{{\mathrm{cosec}}^{2}A}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$: (D)

(i) (B) (ii) (C) (iii) (D) (iv) (D)

$$\text{LHS}={\left(\frac{1-\cos \theta }{\sin \theta }\right)}^2=\frac{(1-\cos \theta {)}^2}{1-{\cos }^2\theta }=\frac{(1-\cos \theta {)}^2}{(1-\cos \theta )(1+\cos \theta )}=\frac{1-\cos \theta }{1+\cos \theta }=\text{RHS}$$

Proved.

$$\text{LHS}=\frac{{\cos }^{2}A+(1+\sin A{)}^2}{(1+\sin A)\cos A}=\frac{{\cos }^{2}A+{\sin }^{2}A+1+2\sin A}{(1+\sin A)\cos A}=\frac{2(1+\sin A)}{(1+\sin A)\cos A}=2\sec A$$

Proved.

Write $t=\tan \theta$, so $\cot \theta =\frac{1}{t}$:

$$\frac{t}{1-\frac{1}{t}}+\frac{\frac{1}{t}}{1-t}=\frac{t^2}{t-1}-\frac{1}{t(t-1)}=\frac{t^3-1}{t(t-1)}=\frac{t^2+t+1}{t}=t+1+\frac{1}{t}$$

Now $\tan \theta +\cot \theta =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\sec \theta \mathrm{cosec}\theta$, so LHS $=1+\sec \theta \mathrm{cosec}\theta$.

Proved.

• LHS $=\frac{1}{\sec A}+1=\cos A+1$
• RHS $=\frac{1-{\cos }^{2}A}{1-\cos A}=\frac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A$

Both sides equal 1 + cos A. Proved.

Divide the numerator and the denominator by $\sin A$:

$$\text{LHS}=\frac{\cot A-1+\mathrm{cosec}A}{\cot A+1-\mathrm{cosec}A}$$

In the numerator, replace $1$ by ${\mathrm{cosec}}^{2}A-{\cot }^{2}A$:

$$\cot A+\mathrm{cosec}A-(\mathrm{cosec}A-\cot A)(\mathrm{cosec}A+\cot A)=(\cot A+\mathrm{cosec}A)(1-\mathrm{cosec}A+\cot A)$$

The second bracket is exactly the denominator, so LHS $=\mathrm{cosec}A+\cot A$.

Proved.

Multiply inside by $\frac{1+\sin A}{1+\sin A}$:

$$\sqrt{\frac{(1+\sin A{)}^2}{1-{\sin }^{2}A}}=\frac{1+\sin A}{\cos A}=\sec A+\tan A$$

Proved.

$$\text{LHS}=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)}=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta [2(1-{\sin }^2\theta )-1]}=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (1-2{\sin }^2\theta )}=\tan \theta$$

Proved.

$$\text{LHS}={\sin }^{2}A+{\mathrm{cosec}}^{2}A+2+{\cos }^{2}A+{\sec }^{2}A+2=5+{\mathrm{cosec}}^{2}A+{\sec }^{2}A$$

$$=5+(1+{\cot }^{2}A)+(1+{\tan }^{2}A)=7+{\tan }^{2}A+{\cot }^{2}A$$

Proved.

• LHS $=\frac{1-{\sin }^{2}A}{\sin A}\cdot \frac{1-{\cos }^{2}A}{\cos A}=\frac{{\cos }^{2}A}{\sin A}\cdot \frac{{\sin }^{2}A}{\cos A}=\sin A\cos A$
• RHS $=\frac{1}{\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}}=\sin A\cos A$

Both sides equal sin A cos A. Proved.

• $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=\frac{{\sec }^{2}A}{{\mathrm{cosec}}^{2}A}={\tan }^{2}A$
• ${\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)}^2={\left(\frac{(1-\tan A)\tan A}{\tan A-1}\right)}^2=(-\tan A{)}^2={\tan }^{2}A$

All three expressions equal tan² A. Proved.