Chapter 8: Introduction to Trigonometry

Step 1 – Interpret $$\tan A = \dfrac{4}{3}$$ geometrically

For an acute angle $$A$$, take a right-angled triangle $$\triangle ABC$$ with right angle at $$B$$. By definition

$$\tan A = \dfrac{\text{side opposite }A}{\text{side adjacent to }A} = \dfrac{BC}{AB}$$

We are given $$\tan A = \dfrac{4}{3}$$, so choose

• $$BC = 4k$$ (opposite)
• $$AB = 3k$$ (adjacent)

for some positive scale factor $$k$$.

Step 2 – Find the hypotenuse $$AC$$ using Pythagoras

In right-angled $$\triangle ABC$$,

$$AC^{2} = AB^{2} + BC^{2}$$

Substitute the chosen lengths:

$$AC^{2} = (3k)^{2} + (4k)^{2} = 9k^{2} + 16k^{2} = 25k^{2}$$

$$\Rightarrow\; AC = \sqrt{25k^{2}} = 5k$$ (hypotenuse).

Step 3 – Calculate all six trigonometric ratios

All required trigonometric ratios of the angle $$A$$ have now been obtained.

Given: $$\angle B$$ and $$\angle Q$$ are acute angles, that is $$0^{\circ} \lt \angle B \lt 90^{\circ}$$ and $$0^{\circ} \lt \angle Q \lt 90^{\circ}$$, and $$\sin B = \sin Q$$.

To prove: $$\angle B = \angle Q$$.

Construction

• Draw a circle with centre $$O$$ and radius $$1$$ (a unit circle).
• Take the right-hand horizontal radius $$OA$$ as the initial side (along the positive $$x$$-axis).
• Mark a point $$P$$ on the circle such that $$\angle AOP = \angle B$$.
• Mark a point $$R$$ on the circle such that $$\angle AOR = \angle Q$$.
• From $$P$$ and $$R$$ drop perpendiculars $$PM$$ and $$RN$$ on $$OA$$ (so $$M$$ and $$N$$ lie on $$OA$$).

Reasoning inside the unit circle

• Because each radius equals $$1$$, in the right-angled triangles $$\triangle OPM$$ and $$\triangle ORN$$ we have $$PM = \sin B$$ and $$RN = \sin Q$$  (opposite side divided by hypotenuse).
• The points $$P$$ and $$R$$ lie in the first quadrant, so $$PM$$ and $$RN$$ equal their vertical ($$y$$-)coordinates.
• As the radius is rotated upward from $$OA$$, its end-point is lifted higher. Therefore:
  if $$\angle B \gt \angle Q$$, then $$P$$ is higher than $$R$$, giving $$PM \gt RN$$;
  if $$\angle B \lt \angle Q$$, then $$P$$ is lower than $$R$$, giving $$PM \lt RN$$.

Proof

Assume, for the sake of contradiction, that $$\angle B \neq \angle Q$$.

• If $$\angle B \gt \angle Q$$, then $$PM \gt RN$$, i.e. $$\sin B \gt \sin Q$$, contradicting the given equality.
• If $$\angle B \lt \angle Q$$, then $$PM \lt RN$$, i.e. $$\sin B \lt \sin Q$$, again contradicting the given equality.

Both alternatives are impossible, so the assumption is wrong. Therefore $$\angle B = \angle Q$$.

Hence proved.

Given : In $$\triangle ACB$$, $$\angle C = 90^{\circ}$$, $$AB = 29\;\text{units}$$, $$BC = 21\;\text{units}$$ and $$\angle ABC = \theta$$.

Step 1 · Find AC

Since the triangle is right-angled at C, by Pythagoras’ theorem

$$AB^{2} = AC^{2} + BC^{2}$$

$$AC^{2} = AB^{2} - BC^{2} = 29^{2} - 21^{2} = 841 - 441 = 400$$

$$\therefore\; AC = \sqrt{400} = 20\;\text{units}$$

Step 2 · Write $$\sin\theta$$ and $$\cos\theta$$

For $$\angle ABC = \theta$$:

• Adjacent side = $$BC = 21$$
• Opposite side = $$AC = 20$$
• Hypotenuse = $$AB = 29$$

Hence

$$\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{21}{29},\qquad \sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{20}{29}$$

Step 3 · Compute the required expressions

1. $$\cos^{2}\theta + \sin^{2}\theta = \left(\dfrac{21}{29}\right)^{2} + \left(\dfrac{20}{29}\right)^{2}$$
   $$= \dfrac{441}{841} + \dfrac{400}{841} = \dfrac{841}{841} = 1$$

2. $$\cos^{2}\theta - \sin^{2}\theta = \left(\dfrac{21}{29}\right)^{2} - \left(\dfrac{20}{29}\right)^{2}$$
   $$= \dfrac{441}{841} - \dfrac{400}{841} = \dfrac{41}{841}$$

Result : $$\cos^{2}\theta + \sin^{2}\theta = 1$$ and $$\cos^{2}\theta - \sin^{2}\theta = \dfrac{41}{841}$$.

Let triangle ABC be right-angled at B, that is, $$\angle B = 90^\circ$$. The acute angle at A satisfies $$\tan A = 1$$.

Step 1 – Express the three sides in a convenient ratio.
For the angle A, $$\tan A = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{BC}{AB} = 1$$
Therefore $$BC = AB$$. Take

• $$AB = x$$
• $$BC = x$$

Using the Pythagoras theorem in right △ABC,

$$AC^2 = AB^2 + BC^2 = x^2 + x^2 = 2x^2 \;\;\Rightarrow\;\; AC = x\sqrt{2}$$

Step 2 – Find $$\sin A$$ and $$\cos A$$.

$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}}$$

$$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Step 3 – Verify the required relation.

$$2 \sin A \cos A = 2 \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = 2 \left(\frac{1}{2}\right) = 1$$

Hence $$2 \sin A \cos A = 1$$, as required.

Step 1 : Assign variables
Let $$PQ = x\,\text{cm}$$. The hypotenuse is then $$OQ = x + 1\,\text{cm}$$ because we are told that $$OQ - PQ = 1\,\text{cm}$$.
The second perpendicular side is given: $$OP = 7\,\text{cm}$$.

Step 2 : Use the Pythagoras theorem
Since the right angle is at P, the Pythagoras theorem gives
$$OP^2 + PQ^2 = OQ^2$$

Substitute the chosen symbols:

$$7^2 + x^2 = (x + 1)^2$$

Simplify:

• $$49 + x^2 = x^2 + 2x + 1$$
• Cancel $$x^2$$ from both sides: $$49 = 2x + 1$$
• $$2x = 48$$
• $$x = 24$$

Thus
$$PQ = 24\,\text{cm}, \qquad OQ = 24 + 1 = 25\,\text{cm}$$.

Step 3 : Identify the sides for angle Q
For angle Q in $$\triangle OPQ$$:

• Opposite side = $$OP = 7\,\text{cm}$$
• Adjacent side = $$PQ = 24\,\text{cm}$$ (the leg that meets Q but is not the hypotenuse)
• Hypotenuse = $$OQ = 25\,\text{cm}$$

Step 4 : Compute $$\sin Q$$ and $$\cos Q$$

$$\sin Q = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$$

$$\cos Q = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$$

Both fractions are already in their lowest terms.

Draw the right triangle $$\triangle PQR$$ with $$\angle Q = 90^{\circ}$$. (In Fig. 8.13, the given lengths are $$PQ = 12\text{ cm}$$ and $$PR = 13\text{ cm}$$.)

Step 1 – Find QR using the Pythagoras theorem
$$PR^{2} = PQ^{2} + QR^{2}$$
$$\Rightarrow QR^{2} = PR^{2} - PQ^{2} = 13^{2} - 12^{2} = 169 - 144 = 25$$
$$\therefore\; QR = 5\text{ cm}.$$

Step 2 – Evaluate $$\tan P$$
For angle $$P$$:

• Opposite side = $$QR = 5\text{ cm}$$
• Adjacent side = $$PQ = 12\text{ cm}$$

Step 3 – Evaluate $$\cot R$$
For angle $$R$$:

• Adjacent side = $$QR = 5\text{ cm}$$
• Opposite side = $$PQ = 12\text{ cm}$$

Step 4 – Compute $$\tan P - \cot R$$
$$\tan P - \cot R = \dfrac{5}{12} - \dfrac{5}{12} = 0.$$

Therefore, $$\tan P - \cot R = 0$$.

Let the given acute angle be $$A$$.

We know the Pythagorean identity

$$\sin^{2}A+\cos^{2}A=1$$

Substitute the given value $$\sin A=\dfrac{3}{4}$$:

$$\cos^{2}A=1-\sin^{2}A$$
$$\cos^{2}A=1-\left(\dfrac{3}{4}\right)^{2}$$
$$\cos^{2}A=1-\dfrac{9}{16}$$
$$\cos^{2}A=\dfrac{16}{16}-\dfrac{9}{16}=\dfrac{7}{16}$$

Taking the positive square-root (because the angle is acute, so its cosine is positive):

$$\cos A=\sqrt{\dfrac{7}{16}}=\dfrac{\sqrt7}{4}$$

Next, use the definition of tangent:

$$\tan A=\dfrac{\sin A}{\cos A}=\dfrac{\dfrac{3}{4}}{\dfrac{\sqrt7}{4}}=\dfrac{3}{\sqrt7}$$

Rationalising the denominator (optional):

$$\tan A=\dfrac{3}{\sqrt7}\times\dfrac{\sqrt7}{\sqrt7}=\dfrac{3\sqrt7}{7}$$

Thus,

$$\cos A=\dfrac{\sqrt7}{4},\qquad \tan A=\dfrac{3}{\sqrt7}=\dfrac{3\sqrt7}{7}$$

We are given $$15 \cot A = 8$$.

First isolate $$\cot A$$:

$$\cot A = \dfrac{8}{15}$$

Recall that for any acute angle $$A$$ drawn in a right-angled triangle,

• $$\cot A = \dfrac{\text{adjacent side}}{\text{opposite side}}$$

We can therefore choose the sides in the same ratio 8 : 15.

Let

$$\text{adjacent side} = 8k, \;\; \text{opposite side} = 15k,$$

where $$k > 0$$ is any convenient common factor.

The hypotenuse $$h$$ follows from Pythagoras’ theorem:

$$h^2 = (8k)^2 + (15k)^2 = 64k^2 + 225k^2 = 289k^2$$

$$\Rightarrow \; h = \sqrt{289k^2} = 17k$$ (since lengths are positive).

Now compute $$\sin A$$ and $$\sec A$$ using the triangle.

• $$\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{15k}{17k} = \dfrac{15}{17}$$
• $$\sec A = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{17k}{8k} = \dfrac{17}{8}$$

The factor $$k$$ cancels out in both ratios, so the values are unique.

Therefore

$$\sin A = \dfrac{15}{17}, \;\; \sec A = \dfrac{17}{8}. $$

Step 1 : Rewrite the given information

The question gives

$$\sec\theta = \dfrac{13}{12}$$

By definition

$$\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent side}}$$

Hence, choose a right-angled triangle in which

• hypotenuse  = 13 units
• adjacent side (to the angle $$\theta$$)  = 12 units

Because every length is positive, we may assume $$\theta$$ lies in the first quadrant, so every trigonometric ratio will be positive.

Step 2 : Find the third side using the Pythagoras theorem

Let the side opposite $$\theta$$ be $$h$$. Then

$$h^{2} = (\text{hypotenuse})^{2} - (\text{adjacent side})^{2}$$

$$h^{2} = 13^{2} - 12^{2}$$

$$h^{2} = 169 - 144 = 25$$

$$\Rightarrow\; h = \sqrt{25} = 5$$

Therefore, the three sides are 5, 12 and 13.

Step 3 : Write all six trigonometric ratios

Step 4 : Verify with reciprocal identities (optional check)

• $$\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{1}{12/13} = \dfrac{13}{12}$$ ✔
• $$\csc\theta = \dfrac{1}{\sin\theta} = \dfrac{1}{5/13} = \dfrac{13}{5}$$ ✔
• $$\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{1}{5/12} = \dfrac{12}{5}$$ ✔

All ratios are consistent, so the solution is complete.

Given : $$\angle A$$ and $$\angle B$$ are acute, that is $$0^{\circ}<A,B<90^{\circ}$$, and $$\cos A = \cos B$$.

We shall prove that $$\angle A = \angle B$$.

Step 1 – Draw two right triangles
Construct

• $$\triangle PQR$$ right-angled at $$Q$$ such that $$\angle PRQ = A$$, and
• $$\triangle XZY$$ right-angled at $$Y$$ such that $$\angle XZY = B$$.

Step 2 – Make the hypotenuses equal
Choose the hypotenuse lengths equal: $$PR = XZ = 1\text{ unit}$$ (any common length works).

Step 3 – Translate the equality of cosines into side equality
By the definition of cosine in a right triangle,
$$\cos A = \frac{RQ}{PR} = RQ$$ (because $$PR = 1$$) and
$$\cos B = \frac{ZY}{XZ} = ZY$$ (because $$XZ = 1$$).
Since $$\cos A = \cos B$$, we get $$RQ = ZY$$.

Step 4 – Establish triangle congruence
In $$\triangle PQR$$ and $$\triangle XZY$$ we now have

• Hypotenuse $$PR = XZ$$ (chosen equal),
• One side $$RQ = ZY$$ (proved equal),
• Each triangle is right-angled (at $$Q$$ and $$Y$$).

Step 5 – Compare the required angles
Corresponding parts of congruent triangles are equal, therefore
$$\angle PRQ = \angle XZY \;\Rightarrow\; A = B$$.

Conclusion
Since the two acute angles have equal cosines, they must themselves be equal.
Hence, $$\boxed{\angle A = \angle B}$$.

We are given $$3\cot A = 4$$.

Therefore
$$\cot A = \frac{4}{3}\;\Rightarrow\;\tan A = \frac{1}{\cot A}=\frac{3}{4}.$$

Left-hand side (L.H.S.)

$$\frac{1-\tan^{2}A}{1+\tan^{2}A}=\frac{1-\left(\frac34\right)^{2}}{1+\left(\frac34\right)^{2}}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}.$$

Right-hand side (R.H.S.)

To find $$\cos^2A-\sin^2A$$ we first determine $$\sin A$$ and $$\cos A$$ from $$\tan A = \frac34$$.

Take a right-angled triangle with sides about $$\angle A$$ as: adjacent = 4, opposite = 3, hypotenuse = $$\sqrt{3^{2}+4^{2}}=5$$.

Thus $$\sin A = \frac35,\; \cos A = \frac45.$$

Hence
$$\cos^{2}A-\sin^{2}A = \left(\frac45\right)^{2}-\left(\frac35\right)^{2}=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}.$$

Since L.H.S. = R.H.S. = $$\frac{7}{25}$$, the identity is verified.

Let the sides of $$\triangle PQR$$ be named with respect to the given information.
The triangle is right-angled at $$Q$$, so $$PR$$ is the hypotenuse and

$$PR^2 = PQ^2 + QR^2$$

Given $$PQ = 5\text{ cm}$$ and $$PR + QR = 25\text{ cm}$$.

Step 1. Express $$PR$$ in terms of $$QR$$

Put $$QR = x\text{ cm}$$. Then

$$PR = \sqrt{PQ^2 + QR^2} = \sqrt{5^2 + x^2} = \sqrt{25 + x^2}$$

Step 2. Use the additional relation

$$PR + QR = 25 \;\Rightarrow\; \sqrt{25 + x^2} + x = 25$$

Move $$x$$ to the other side:

$$\sqrt{25 + x^2} = 25 - x$$

Square both sides:

$$25 + x^2 = (25 - x)^2 = 625 - 50x + x^2$$

Cancel $$x^2$$ and solve for $$x$$:

$$25 = 625 - 50x \;\Rightarrow\; 50x = 600 \;\Rightarrow\; x = 12$$

Hence $$QR = 12\text{ cm}$$.

Step 3. Find $$PR$$

$$PR = 25 - QR = 25 - 12 = 13\text{ cm}$$

So the three sides are:

• $$PQ = 5\text{ cm}$$ (adjacent to $$\angle P$$)
• $$QR = 12\text{ cm}$$ (opposite to $$\angle P$$)
• $$PR = 13\text{ cm}$$ (hypotenuse)

Step 4. Compute the required trigonometric ratios

$$\sin P = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{12}{13}$$

$$\cos P = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{5}{13}$$

$$\tan P = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{12}{5}$$

All the required values have been determined.

Step 1 : Identify the hypotenuse and the sides with respect to $$\angle ACB$$

The triangle is right-angled at $$B$$, therefore $$AC$$ is the hypotenuse.
For the acute angle $$\angle ACB = 30^{\circ}$$ :
  • the side opposite to the angle is $$AB$$ (given $$5\ \text{cm}$$).
  • the side adjacent to the angle is $$BC$$ (to be found).

Step 2 : Find the hypotenuse $$AC$$ using $$\sin 30^{\circ}$$

By definition, $$\sin 30^{\circ} = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{AB}{AC}$$.

Substituting the known value, $$\dfrac{1}{2} = \dfrac{5}{AC}$$.

Cross-multiplying, $$AC = 5 \times 2 = 10\ \text{cm}$$.

Step 3 : Find the remaining side $$BC$$ using $$\cos 30^{\circ}$$

$$\cos 30^{\circ} = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{BC}{AC}$$.

Substituting $$\cos 30^{\circ} = \dfrac{\sqrt{3}}{2}$$ and $$AC = 10\ \text{cm}$$, we get
$$\dfrac{\sqrt{3}}{2} = \dfrac{BC}{10}$$.

Therefore $$BC = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\ \text{cm}$$.

Step 4 : Write the results

$$BC = 5\sqrt{3}\ \text{cm}, \quad AC = 10\ \text{cm}$$.

In △PQR, Q is the right angle, so $$\angle PQR = 90^{\circ}$$ and $$PR$$ is the hypotenuse.

1. Find the third side $$QR$$.

By Pythagoras’ theorem, $$PR^{2} = PQ^{2} + QR^{2}$$.

$$6^{2} = 3^{2} + QR^{2}$$

$$36 = 9 + QR^{2}$$

$$QR^{2} = 27 \;\Rightarrow\; QR = \sqrt{27} = 3\sqrt{3}\text{ cm}$$.

2. Calculate $$\angle QPR$$ (at P).

Adjacent side = $$PQ$$, hypotenuse = $$PR$$.

$$\cos(\angle QPR) = \frac{PQ}{PR} = \frac{3}{6} = \frac12$$

Hence $$\angle QPR = 60^{\circ}$$ because $$\cos 60^{\circ} = \frac12$$.

3. Calculate $$\angle PRQ$$ (at R).

Adjacent side = $$QR$$, hypotenuse = $$PR$$.

$$\cos(\angle PRQ) = \frac{QR}{PR} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}$$

Hence $$\angle PRQ = 30^{\circ}$$ because $$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$$.

4. Verification.

$$60^{\circ} + 30^{\circ} + 90^{\circ} = 180^{\circ}$$  ✓

Therefore, $$\angle QPR = 60^{\circ}$$ and $$\angle PRQ = 30^{\circ}$$.

We are given:

• $$\sin(A-B) = \dfrac{1}{2}$$
• $$\cos(A+B) = \dfrac{1}{2}$$
• $$0^{\circ} \lt A+B \le 90^{\circ}$$ and $$A \gt B$$

Step 1: Find $$A+B$$ using the cosine value.

For an angle $$\theta$$ with $$0^{\circ} \lt \theta \le 90^{\circ}$$ (the first quadrant), the equation $$\cos\theta = \dfrac{1}{2}$$ has exactly one solution, namely $$\theta = 60^{\circ}$$, because $$\cos 60^{\circ} = \dfrac{1}{2}$$.

Since $$A+B$$ itself lies in this range, we conclude

$$A+B = 60^{\circ}. \quad(1)$$

Step 2: Find $$A-B$$ using the sine value.

Because $$A \gt B$$, the difference $$A-B$$ is positive; and it satisfies $$A-B \lt A+B \le 90^{\circ}$$, so $$A-B$$ also lies in the first quadrant. In that range, $$\sin\phi = \dfrac{1}{2}$$ has the unique solution $$\phi = 30^{\circ}$$ (since $$\sin 30^{\circ} = \dfrac{1}{2}$$). Therefore

$$A-B = 30^{\circ}. \quad(2)$$

Step 3: Solve the two linear equations.

Add (1) and (2):
$$(A+B) + (A-B) = 60^{\circ} + 30^{\circ} \;\Rightarrow\; 2A = 90^{\circ} \;\Rightarrow\; A = 45^{\circ}.$$

Substitute $$A = 45^{\circ}$$ in (1):
$$45^{\circ} + B = 60^{\circ} \;\Rightarrow\; B = 15^{\circ}.$$

Step 4: Verify the conditions.

• $$0^{\circ} \lt A+B = 60^{\circ} \le 90^{\circ}$$  ✓
• $$A = 45^{\circ} \gt 15^{\circ} = B$$  ✓
• $$\sin(A-B) = \sin 30^{\circ} = \dfrac{1}{2}$$  ✓  and  $$\cos(A+B) = \cos 60^{\circ} = \dfrac{1}{2}$$  ✓

Hence the required angles are $$A = 45^{\circ}$$ and $$B = 15^{\circ}$$.

Step 1 : Interpret the first condition

We know that $$\tan 60^{\circ}=\sqrt{3}$$. In general, if $$\tan\theta=\sqrt{3}$$, then $$\theta=60^{\circ}+k\,180^{\circ}$$ for any integer $$k$$.

The question states $$0^{\circ}<A+B\le 90^{\circ}$$, so the only value from the general family that satisfies this range is

$$A+B=60^{\circ} \qquad(1)$$

Step 2 : Interpret the second condition

We know that $$\tan 30^{\circ}=\dfrac1{\sqrt3}$$. Hence, if $$\tan\theta=\dfrac1{\sqrt3}$$, then $$\theta=30^{\circ}+k\,180^{\circ}$$.

The expression $$A-B$$ represents the difference of two angles with $$A>B$$, so $$A-B$$ must be positive and less than $$A+B\le90^{\circ}$$. Choosing $$k=0$$ gives the only admissible value:

$$A-B=30^{\circ} \qquad(2)$$

Step 3 : Solve the simultaneous equations

Add equations (1) and (2):

$$\bigl(A+B\bigr)+\bigl(A-B\bigr)=60^{\circ}+30^{\circ}$$

$$2A=90^{\circ}$$

$$A=45^{\circ}$$

Substitute $$A=45^{\circ}$$ into equation (2):

$$45^{\circ}-B=30^{\circ}$$

$$B=15^{\circ}$$

Step 4 : Verification

• $$A+B=45^{\circ}+15^{\circ}=60^{\circ}$$  ⇒  $$\tan(A+B)=\tan60^{\circ}=\sqrt{3}$$ (matches given).
• $$A-B=45^{\circ}-15^{\circ}=30^{\circ}$$  ⇒  $$\tan(A-B)=\tan30^{\circ}=\dfrac1{\sqrt3}$$ (matches given).
• $$A=45^{\circ}>15^{\circ}=B$$ and $$0^{\circ}<60^{\circ}\le90^{\circ}$$, so all stated conditions are satisfied.

Hence, the required angles are $$A=45^{\circ}$$ and $$B=15^{\circ}$$.

Given: We have to rewrite the trigonometric ratios $$\cos A$$, $$\tan A$$ and $$\sec A$$ only in terms of $$\sin A$$. We take angle $$A$$ to be acute (that is, $$0^{\circ}<A<90^{\circ}$$), so every primary ratio is positive.

Step 1 : Relate $$\cos A$$ to $$\sin A$$

The Pythagorean identity for any angle $$A$$ is

$$\sin^{2}A+\cos^{2}A=1$$

Rearrange to isolate $$\cos^{2}A$$:

$$\cos^{2}A = 1-\sin^{2}A$$

Because $$A$$ is acute, $$\cos A>0$$; therefore,

$$\cos A = \sqrt{1-\sin^{2}A}$$

Step 2 : Express $$\tan A$$ in terms of $$\sin A$$

By definition,

$$\tan A = \frac{\sin A}{\cos A}$$

Substitute the value of $$\cos A$$ found in Step 1:

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

Step 3 : Express $$\sec A$$ in terms of $$\sin A$$

Since $$\sec A = \dfrac{1}{\cos A}$$, substitute $$\cos A=\sqrt{1-\sin^{2}A}$$:

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

Conclusion

All three required ratios have been rewritten solely with $$\sin A$$.

To prove: $$\sec A (1 - \sin A)(\sec A + \tan A) = 1$$.

Step 1 – Write every term in sine and cosine.

$$\sec A = \dfrac{1}{\cos A}, \qquad \tan A = \dfrac{\sin A}{\cos A}$$

Substituting these in the left–hand expression gives

$$\sec A(1 - \sin A)(\sec A + \tan A) = \dfrac{1}{\cos A}(1 - \sin A)\! \left(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\right).$$

Step 2 – Combine the terms inside the last bracket.

$$\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A} = \dfrac{1 + \sin A}{\cos A}.$$

So the expression becomes

$$\dfrac{1}{\cos A}(1 - \sin A)\,\dfrac{1 + \sin A}{\cos A} = \dfrac{(1 - \sin A)(1 + \sin A)}{\cos^{2} A}.$$

Step 3 – Multiply the two conjugates.

$$(1 - \sin A)(1 + \sin A) = 1 - \sin^{2} A.$$

Step 4 – Use the Pythagorean identity.

From $$\sin^{2} A + \cos^{2} A = 1$$, we have $$1 - \sin^{2} A = \cos^{2} A$$.

Therefore,

$$\dfrac{1 - \sin^{2} A}{\cos^{2} A} = \dfrac{\cos^{2} A}{\cos^{2} A} = 1.$$

Thus, $$\sec A(1 - \sin A)(\sec A + \tan A) = 1.$$

We have to prove

$$\dfrac{\cot A - \cos A}{\cot A + \cos A} = \dfrac{\operatorname{cosec} A - 1}{\operatorname{cosec} A + 1}.$$

Recall the reciprocal identities

• $$\cot A = \dfrac{\cos A}{\sin A}$$
• $$\operatorname{cosec} A = \dfrac{1}{\sin A}$$

Domain. Throughout the proof we assume that

• $$\sin A \neq 0$$, so that $$\cot A$$ and $$\operatorname{cosec} A$$ (and hence both sides of the identity) are defined; and
• $$\cos A \neq 0$$, so that in Step 1 we may legitimately cancel the common factor $$\cos A$$ from the numerator and the denominator.

Step 1: Rewrite the left-hand side (LHS) in terms of $$\sin A$$ and $$\cos A$$.

$$\cot A - \cos A = \dfrac{\cos A}{\sin A} - \cos A = \cos A\left(\dfrac{1}{\sin A} - 1\right)$$

$$\cot A + \cos A = \dfrac{\cos A}{\sin A} + \cos A = \cos A\left(\dfrac{1}{\sin A} + 1\right)$$

Hence

$$\dfrac{\cot A - \cos A}{\cot A + \cos A} = \dfrac{\cos A\left(\dfrac{1}{\sin A} - 1\right)}{\cos A\left(\dfrac{1}{\sin A} + 1\right)}.$$

Since $$\cos A \neq 0$$, the common factor $$\cos A$$ cancels, leaving

$$\dfrac{\cot A - \cos A}{\cot A + \cos A} = \dfrac{\dfrac{1}{\sin A} - 1}{\dfrac{1}{\sin A} + 1}.$$

Step 2: Clear the complex fractions (multiply numerator and denominator by $$\sin A$$).

$$\dfrac{\dfrac{1}{\sin A} - 1}{\dfrac{1}{\sin A} + 1} = \dfrac{1 - \sin A}{1 + \sin A}.$$

So the LHS equals $$\dfrac{1 - \sin A}{1 + \sin A}$$.

Step 3: Simplify the right-hand side (RHS).

Using $$\operatorname{cosec} A = \dfrac{1}{\sin A}$$ we get

$$\dfrac{\operatorname{cosec} A - 1}{\operatorname{cosec} A + 1} = \dfrac{\dfrac{1}{\sin A} - 1}{\dfrac{1}{\sin A} + 1}.$$

Multiplying numerator and denominator by $$\sin A$$:

$$\dfrac{\operatorname{cosec} A - 1}{\operatorname{cosec} A + 1} = \dfrac{1 - \sin A}{1 + \sin A}.$$

Step 4: Compare the two expressions.

Both LHS and RHS reduce to the same expression $$\dfrac{1 - \sin A}{1 + \sin A}$$. Therefore

$$\dfrac{\cot A - \cos A}{\cot A + \cos A} = \dfrac{\operatorname{cosec} A - 1}{\operatorname{cosec} A + 1}.$$

Hence proved.

Let
$$LHS = \dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} \quad\text{and}\quad RHS = \dfrac{1}{\sec\theta - \tan\theta}.$$

Step 1   Simplify the right-hand side

Rationalise the denominator of $$RHS$$:

$$\dfrac{1}{\sec\theta-\tan\theta}\;=\;\dfrac{1}{\sec\theta-\tan\theta}\,\cdot\,\dfrac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}\;= \;\dfrac{\sec\theta+\tan\theta}{\sec^{2}\theta-\tan^{2}\theta}.$$

Using $$\sec^{2}\theta = 1+\tan^{2}\theta$$, we get

$$\sec^{2}\theta-\tan^{2}\theta=(1+\tan^{2}\theta)-\tan^{2}\theta=1.$$

Therefore

$$RHS = \sec\theta+\tan\theta.$$

Express each term in sine and cosine:

$$\sec\theta+\tan\theta=\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}=\dfrac{1+\sin\theta}{\cos\theta}.$$ Hence

$$RHS = \dfrac{1+\sin\theta}{\cos\theta}. \quad(1)$$

Step 2   Show that the left-hand side equals $$(1)$$

Let $$s=\sin\theta$$ and $$c=\cos\theta$$ (to lighten notation).
We need to prove

$$\dfrac{s-c+1}{s+c-1}=\dfrac{1+s}{c}. \quad(2)$$

Cross-multiply the two fractions in (2):

$$c\,(s-c+1)=(1+s)(s+c-1).$$

Compute each side separately.

• Left side:
  $$c\,(s-c+1)=cs-c^{2}+c.$$
• Right side:
  $$(1+s)(s+c-1)=1\cdot(s+c-1)+s\cdot(s+c-1) \\ \qquad = (s+c-1)+(s^{2}+sc-s) \\ \qquad = s^{2}+sc+c-1.$$

Take the difference “right − left”:

$$\bigl(s^{2}+sc+c-1\bigr)-\bigl(cs-c^{2}+c\bigr)=s^{2}+sc+c-1-cs+c^{2}-c=s^{2}+c^{2}-1.$$

But for every angle $$\theta$$, $$s^{2}+c^{2}=\sin^{2}\theta+\cos^{2}\theta=1.$$ Hence the difference is $$0$$, so the two sides are equal. Therefore equation (2) is true, i.e.

$$\dfrac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\dfrac{1+\sin\theta}{\cos\theta}. \quad(3)$$

Step 3   Conclude the proof

From (1) and (3) we have

$$LHS=\dfrac{1+\sin\theta}{\cos\theta}=RHS.$$

Hence the required identity is proved:

$$\boxed{\dfrac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\dfrac{1}{\sec\theta-\tan\theta}}.$$

The given task is to rewrite the basic ratios $$\sin A$$, $$\sec A$$ and $$\tan A$$ only through $$\cot A$$.

1. Expressing $$\tan A$$

By definition $$\cot A$$ is the reciprocal of $$\tan A$$, therefore

$$\cot A = \dfrac{1}{\tan A} \;\;\Longrightarrow\;\; \tan A = \dfrac{1}{\cot A}.$$

2. Expressing $$\sin A$$

For any angle A we have the Pythagorean identity

$$1 + \cot^{2} A = \cosec^{2} A.$$

Taking the positive square root (angles dealt with in Class 10 are acute, so every trigonometric value is positive), we get

$$\cosec A = \sqrt{1 + \cot^{2} A}.$$

Since $$\sin A$$ is the reciprocal of $$\cosec A$$,

$$\sin A = \dfrac{1}{\cosec A} = \dfrac{1}{\sqrt{1 + \cot^{2} A}}.$$

3. Expressing $$\sec A$$

Begin with the identity

$$\sec^{2} A = 1 + \tan^{2} A.$$

Substitute $$\tan A = \dfrac{1}{\cot A}$$ obtained in step 1:

$$\sec^{2} A = 1 + \left(\dfrac{1}{\cot A}\right)^{2} = 1 + \dfrac{1}{\cot^{2} A} = \dfrac{\cot^{2} A + 1}{\cot^{2} A}.$$

Again taking the positive square root,

$$\sec A = \sqrt{\dfrac{\cot^{2} A + 1}{\cot^{2} A}} = \dfrac{\sqrt{1 + \cot^{2} A}}{\cot A}.$$

4. Collected results

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

Let us denote $$\sec A = k$$  (where $$k > 1$$ for an acute angle).

1. Express $$\cos A$$ in terms of $$\sec A$$
By definition, $$\sec A = \dfrac{1}{\cos A}$$, therefore
$$\cos A = \dfrac{1}{\sec A}$$.

2. Obtain $$\sin A$$ using the Pythagorean identity
We know $$\sin^{2}A + \cos^{2}A = 1$$. Substitute $$\cos A = \dfrac{1}{\sec A}$$:

$$\sin^{2}A + \left(\dfrac{1}{\sec A}\right)^{2} = 1$$

$$\sin^{2}A + \dfrac{1}{\sec^{2}A} = 1$$

$$\sin^{2}A = 1 - \dfrac{1}{\sec^{2}A}$$

$$\sin^{2}A = \dfrac{\sec^{2}A - 1}{\sec^{2}A}$$

Taking the positive square root (for an acute angle),
$$\sin A = \dfrac{\sqrt{\sec^{2}A - 1}}{\sec A}$$.

3. Derive the remaining ratios

• Tangent
  $$\tan A = \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{\sqrt{\sec^{2}A - 1}}{\sec A}}{\dfrac{1}{\sec A}} = \sqrt{\sec^{2}A - 1}$$
• Cotangent
  $$\cot A = \dfrac{1}{\tan A} = \dfrac{1}{\sqrt{\sec^{2}A - 1}}$$
• Cosecant
  $$\csc A = \dfrac{1}{\sin A} = \dfrac{1}{\dfrac{\sqrt{\sec^{2}A - 1}}{\sec A}} = \dfrac{\sec A}{\sqrt{\sec^{2}A - 1}}$$

4. Summary