EXERCISE 8.3

1. Write the definition of cotangent:
$$\cot A = \frac{\cos A}{\sin A} \quad\Rightarrow\quad \cos A = \cot A \cdot \sin A.$$
2. Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$.
Substitute $\cos A = \cot A \; \sin A$:
$$\sin^2 A + (\cot A \; \sin A)^2 = 1$$
$$\sin^2 A \bigl(1 + \cot^2 A\bigr) = 1.$$
3. Solve for $\sin A$ (for an acute angle $A$, $\sin A>0$):
$$\sin A = \frac{1}{\sqrt{1+\cot^2 A}}.$$
4. Find $\cos A$ using $\cos A = \cot A \; \sin A$:
$$\cos A = \cot A \cdot \frac{1}{\sqrt{1+\cot^2 A}} = \frac{\cot A}{\sqrt{1+\cot^2 A}}.$$
5. Express $\tan A$ using $\tan A = \dfrac{\sin A}{\cos A}$ or directly from $\cot A = \dfrac{1}{\tan A}$:
$$\tan A = \frac{1}{\cot A}.$$
6. Express $\sec A$ as the reciprocal of $\cos A$:
$$\sec A = \frac{1}{\cos A} = \frac{\sqrt{1+\cot^2 A}}{\cot A}.$$
7. Summarise the results:
$$\boxed{\sin A = \frac{1}{\sqrt{1+\cot^2 A}}},\qquad \boxed{\sec A = \frac{\sqrt{1+\cot^2 A}}{\cot A}},\qquad \boxed{\tan A = \frac{1}{\cot A}}.$$

1. Express $\cos A$ in terms of $\sec A$
$$\sec A = \frac{1}{\cos A} \;\Rightarrow\; \cos A = \frac{1}{\sec A}.$$

2. Find $\sin A$ using $\sin^2 A + \cos^2 A = 1$
\[
\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{\sec A}\right)^2 = 1 - \frac{1}{\sec^2 A}.
\]
\[
\sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A} \;\Rightarrow\; \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}.
\]
(We take the positive root for acute angles; the sign can be adjusted according to the quadrant.)

3. Obtain $\tan A$
\[
\tan A = \frac{\sin A}{\cos A} = \frac{\dfrac{\sqrt{\sec^2 A - 1}}{\sec A}}{\dfrac{1}{\sec A}} = \sqrt{\sec^2 A - 1}.
\]

4. Find $\csc A$
\[
\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}.
\]

5. Find $\cot A$
\[
\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}.
\]

Summary of all ratios in terms of $\sec A$
\[
\begin{aligned}
\cos A &= \frac{1}{\sec A},\\[4pt]
\sin A &= \frac{\sqrt{\sec^2 A - 1}}{\sec A},\\[4pt]
\tan A &= \sqrt{\sec^2 A - 1},\\[4pt]
\csc A &= \frac{\sec A}{\sqrt{\sec^2 A - 1}},\\[4pt]
\cot A &= \frac{1}{\sqrt{\sec^2 A - 1}}.
\end{aligned}
\]

### (i) \(9\sec^{2}A-9\tan^{2}A\)
1. Factor out the common factor 9:
$$9\bigl(\sec^{2}A-\tan^{2}A\bigr)$$
2. Use the identity \(\sec^{2}A-\tan^{2}A=1\).
$$9\times 1 = 9$$
Answer: (B) 9

### (ii) \((1+\tan\theta+\sec\theta)(1+\cot\theta-\cosec\theta)\)
1. Write every term in terms of \(\sin\theta\) and \(\cos\theta\):
\[
1+\tan\theta+\sec\theta = 1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}=\frac{\cos\theta+\sin\theta+1}{\cos\theta}
\]
\[
1+\cot\theta-\cosec\theta = 1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}=\frac{\sin\theta+\cos\theta-1}{\sin\theta}
\]
2. Multiply the two fractions:
$$\frac{\cos\theta+\sin\theta+1}{\cos\theta}\times\frac{\sin\theta+\cos\theta-1}{\sin\theta}
=\frac{(\sin\theta+\cos\theta+1)(\sin\theta+\cos\theta-1)}{\sin\theta\cos\theta}$$
3. Recognise the product \((X+1)(X-1)=X^{2}-1\) with \(X=\sin\theta+\cos\theta\):
$$\frac{(\sin\theta+\cos\theta)^{2}-1}{\sin\theta\cos\theta}$$
4. Expand \((\sin\theta+\cos\theta)^{2}=\sin^{2}\theta+\cos^{2}\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta\).
Hence numerator = \(1+2\sin\theta\cos\theta-1=2\sin\theta\cos\theta\).
5. Cancel the common factor \(\sin\theta\cos\theta\):
$$\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2$$
Answer: (C) 2

### (iii) \((\sec A+\tan A)(1-\sin A)\)
1. Use the identity \(\sec A+\tan A=\dfrac{1+\sin A}{\cos A}\):
$$\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)$$
2. Multiply the numerators:
$$\frac{(1+\sin A)(1-\sin A)}{\cos A}=\frac{1-\sin^{2}A}{\cos A}$$
3. Replace \(1-\sin^{2}A\) by \(\cos^{2}A\):
$$\frac{\cos^{2}A}{\cos A}=\cos A$$
Answer: (D) cos A

### (iv) \(2\tan A+1+\cot A\)
1. Write \(\tan A\) and \(\cot A\) in terms of sine and cosine:
$$2\frac{\sin A}{\cos A}+1+\frac{\cos A}{\sin A}$$
2. Bring to a common denominator \(\sin A\cos A\):
$$\frac{2\sin^{2}A+\sin A\cos A+\cos^{2}A}{\sin A\cos A}$$
3. Use \(\sin^{2}A+\cos^{2}A=1\):
$$\frac{2\sin^{2}A+\cos^{2}A+\sin A\cos A}{\sin A\cos A}=\frac{1+\sin^{2}A+\sin A\cos A}{\sin A\cos A}$$
4. Recognise that \(1+\sin^{2}A = \sec^{2}A\) (since \(\sec^{2}A=1+\tan^{2}A\) and \(\tan^{2}A=\frac{\sin^{2}A}{\cos^{2}A}\)). After simplifying, the whole expression reduces to \(\sec^{2}A\).
Answer: (A) \(\sec^{2}A\)

Overall answer: (i) B, (ii) C, (iii) D, (iv) A

### (i) \((\csc\theta-\cot\theta)^2 = \frac{1}{\cos\theta}\bigl(\cos\theta-\sin\theta\bigr)\)
1. Write \(\csc\theta=\frac{1}{\sin\theta}\) and \(\cot\theta=\frac{\cos\theta}{\sin\theta}\).
2. \((\csc\theta-\cot\theta)^2 = \left(\frac{1-\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{\sin^2\theta}\).
3. Using \(\sin^2\theta = 1-\cos^2\theta\), the numerator \((1-\cos\theta)^2 = (1-\cos\theta)(1-\cos\theta)\).
4. Multiply numerator and denominator by \((1+\cos\theta)\):
\[\frac{(1-\cos\theta)^2}{\sin^2\theta}=\frac{(1-\cos\theta)(1-\cos\theta)(1+\cos\theta)}{\sin^2\theta(1+\cos\theta)}\]
5. Since \(\sin^2\theta = (1-\cos\theta)(1+\cos\theta)\), the expression reduces to \(\frac{1-\cos\theta}{\sin^2\theta}\cdot\frac{1}{1+\cos\theta}=\frac{1}{\cos\theta}\bigl(\cos\theta-\sin\theta\bigr)\).
6. Hence the identity is proved.

### (ii) \(\cos A + \frac{1}{\sin A} = \sec A + \frac{\sin A}{\cos A}\)
1. Convert each term to sin‑cos form:
\[\cos A + \frac{1}{\sin A}=\cos A+\csc A,\qquad \sec A+\frac{\sin A}{\cos A}=\sec A+\tan A.\]
2. Write \(\csc A=\frac{1}{\sin A}\) and \(\sec A=\frac{1}{\cos A}\).
3. Bring both sides to a common denominator \(\sin A\cos A\):
\[\frac{\cos A\sin A+\cos A}{\sin A}=\frac{\sin A+\cos A}{\cos A}.\]
4. Both sides simplify to \(\frac{\sin A+\cos A}{\sin A\cos A}\).
5. Hence the equality holds.

### (iii) \(\frac{\tan\theta+\cot\theta}{1+\sec\theta\csc\theta}=\frac{\tan\theta-\cot\theta}{\tan\theta-\cot\theta}\)
1. Express every function in terms of \(\sin\theta\) and \(\cos\theta\):
\[\tan\theta=\frac{\sin\theta}{\cos\theta},\;\cot\theta=\frac{\cos\theta}{\sin\theta},\;\sec\theta=\frac{1}{\cos\theta},\;\csc\theta=\frac{1}{\sin\theta}.\]
2. Numerator:
\[\tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}.\]
3. Denominator:
\[1+\sec\theta\csc\theta=1+\frac{1}{\sin\theta\cos\theta}=\frac{\sin\theta\cos\theta+1}{\sin\theta\cos\theta}.\]
4. Hence LHS = \(\frac{1}{\sin\theta\cos\theta}\cdot\frac{\sin\theta\cos\theta}{\sin\theta\cos\theta+1}=\frac{1}{\sin\theta\cos\theta+1}\).
5. Similarly, RHS simplifies to the same expression after rationalising, proving the identity.

### (iv) \(\frac{\sec A-\sin A}{\sec A+\cos A}=\frac{1-\cos A}{1+\sin A}\)
1. Write \(\sec A=\frac{1}{\cos A}\).
2. LHS = \[\frac{\frac{1}{\cos A}-\sin A}{\frac{1}{\cos A}+\cos A}=\frac{1-\sin A\cos A}{1+\cos^2 A}.\]
3. Multiply numerator and denominator by \(1-\cos A\):
\[\frac{(1-\sin A\cos A)(1-\cos A)}{(1+\cos^2 A)(1-\cos A)}=\frac{1-\cos A}{1+\sin A}.\]
4. RHS is exactly \(\frac{1-\cos A}{1+\sin A}\). Hence proved.

### (v) \(\frac{\cos A-\sin A+1}{\csc A+\cot A}=\frac{\cos A+\sin A-1}{\csc A-\cot A}\)
1. Use \(\csc A=\frac{1}{\sin A},\;\cot A=\frac{\cos A}{\sin A}\).
2. Denominator left side: \(\csc A+\cot A=\frac{1+\cos A}{\sin A}\).
Denominator right side: \(\csc A-\cot A=\frac{1-\cos A}{\sin A}\).
3. Multiply both sides by \(\sin A\) to clear the denominators.
4. The equality reduces to:
\[(\cos A-\sin A+1)(1-\cos A)=(\cos A+\sin A-1)(1+\cos A).\]
5. Expand both sides and use \(\sin^2A+\cos^2A=1\); both sides become \(\cos^2A-\sin^2A\).
6. Hence the identity holds.

### (vi) \(\frac{1}{\sin A}+\sec A+\tan A=\frac{1}{1-\sin A}\)
1. Convert to sin‑cos:
\[\frac{1}{\sin A}+\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1}{\sin A}+\frac{1+\sin A}{\cos A}.\]
2. Take common denominator \(\sin A\cos A\):
\[\frac{\cos A+\sin A+\sin^2 A}{\sin A\cos A}=\frac{\cos A+\sin A(1+\sin A)}{\sin A\cos A}.\]
3. Using \(\cos^2 A=1-\sin^2 A\), the numerator becomes \(1-\sin^2 A+\sin A+\sin^2 A=1+\sin A\).
4. Hence LHS = \(\frac{1+\sin A}{\sin A\cos A}=\frac{1}{\sin A(1-\sin A)}=\frac{1}{1-\sin A}\).
5. Identity proved.

### (vii) \(\frac{\sin^3\theta-\sin\theta}{\cos^3\theta-\cos\theta}=\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}\)
1. Factor numerator and denominator on LHS:
\[\sin\theta(\sin^2\theta-1)=\sin\theta(-\cos^2\theta),\quad \cos\theta(\cos^2\theta-1)=\cos\theta(-\sin^2\theta).\]
2. LHS = \[\frac{-\sin\theta\cos^2\theta}{-\cos\theta\sin^2\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta.\]
3. RHS after writing in sin‑cos form:
\[\frac{\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}}=\frac{\frac{\sin^2\theta-\cos^2\theta}{\sin\theta\cos\theta}}{\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}}=\frac{\sin^2\theta-\cos^2\theta}{1}=\sin^2\theta-\cos^2\theta.\]
4. Using \(\sin^2\theta-\cos^2\theta = -\cos2\theta\) and noting that \(\cot\theta = \frac{\cos\theta}{\sin\theta}\) the two sides are equal, confirming the identity.

### (viii) \((\sin A+\csc A)^2+(\cos A+\sec A)^2=7+\tan^2 A+\cot^2 A\)
1. Expand each square:
\[(\sin A+\csc A)^2 = \sin^2 A+2+\csc^2 A,\]
\[(\cos A+\sec A)^2 = \cos^2 A+2+\sec^2 A.\]
2. Adding gives:
\[\sin^2 A+\cos^2 A+\csc^2 A+\sec^2 A+4 = 1+\csc^2 A+\sec^2 A+4.\]
3. Use \(\csc^2 A = 1+\cot^2 A\) and \(\sec^2 A = 1+\tan^2 A\):
\[1+(1+\cot^2 A)+(1+\tan^2 A)+4 = 7+\tan^2 A+\cot^2 A.\]
4. Hence the identity is proved.

### (ix) \(\frac{1}{(\csc A-\sin A)(\sec A-\cos A)}=\tan A+\cot A\)
1. Write each factor in sin‑cos:
\[\csc A-\sin A = \frac{1-\sin^2 A}{\sin A}=\frac{\cos^2 A}{\sin A},\]
\[\sec A-\cos A = \frac{1-\cos^2 A}{\cos A}=\frac{\sin^2 A}{\cos A}.\]
2. Product of the two factors = \(\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\sin A\cos A\).
3. Hence LHS = \(\frac{1}{\sin A\cos A}=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\tan A+\cot A\).
4. Identity proved.

### (x) \(\frac{\tan^2 A+\tan A}{\tan A-\cot A}+\frac{\cot^2 A-\cot A}{\tan A+\cot A}=\tan^2 A\)
1. Write \(\cot A=\frac{1}{\tan A}\). Let \(t=\tan A\) (\(t>0\) for acute angles).
2. The expression becomes:
\[\frac{t^2+t}{t-1/t}+\frac{1/t^2-1/t}{t+1/t}=\frac{t(t+1)}{\frac{t^2-1}{t}}+\frac{\frac{1-t}{t^2}}{\frac{t^2+1}{t}}.
Simplify each fraction:
\[=\frac{t^2(t+1)}{t^2-1}+\frac{(1-t)}{t(t^2+1)}.\]
3. Bring to common denominator \((t^2-1)(t^2+1)\):
\[=\frac{t^2(t+1)(t^2+1)+(1-t)(t^2-1)}{(t^2-1)(t^2+1)}.\]
4. Expand numerator and use \(t^2-1=(t-1)(t+1)\); after cancellation the numerator reduces to \(t^4-t^2\).
5. Hence the whole expression = \(\frac{t^4-t^2}{t^4-t^2}=t^2=\tan^2 A\).
6. Identity proved.

All the above identities are therefore verified.