Question 4

Acute angles: All angles here are acute (between 0° and 90°), so all trig ratios are positive. This allows us to cancel common factors and take positive square roots without worrying about sign changes.

Proof strategy: We simplify one side (usually LHS) by converting everything to $\sin$ and $\cos$, then simplify until it matches the other side.

We will prove each identity by simplifying one or both sides until they are equal.

Step 1 — Proving identity (i)

Let's start with the left-hand side (LHS) of the identity. We will convert cosec $\theta$ and cot $\theta$ into terms of $\sin \theta$ and $\cos \theta$.

$(\text{cosec } \theta - \cot \theta)^2$

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

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

$= \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$

Pythagorean Identity: $\sin^2\theta = 1 - \cos^2\theta$.

Difference of Squares: $1 - \cos^2\theta = (1-\cos\theta)(1+\cos\theta)$.

We know that $\sin^2 \theta = 1 - \cos^2 \theta$.

$= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$

We can factor the denominator as a difference of squares: $1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$.

$= \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$

We can cancel out the common term $(1 - \cos \theta)$.

$= \frac{1 - \cos \theta}{1 + \cos \theta}$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 2 — Proving identity (ii)

Let's start with the left-hand side (LHS) of the identity. We will find a common denominator for the two fractions.

$\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$

$= \frac{\cos A \cdot \cos A + (1 + \sin A) \cdot (1 + \sin A)}{(1 + \sin A) \cos A}$

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

Expand $(1 + \sin A)^2$ as $1 + 2 \sin A + \sin^2 A$.

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

Pythagorean Identity: $\cos^2 A + \sin^2 A = 1$.

Group $\cos^2 A + \sin^2 A$, which equals 1.

$= \frac{1 + 1 + 2 \sin A}{(1 + \sin A) \cos A}$

$= \frac{2 + 2 \sin A}{(1 + \sin A) \cos A}$

Factor out 2 from the numerator.

$= \frac{2(1 + \sin A)}{(1 + \sin A) \cos A}$

Cancel out the common term $(1 + \sin A)$.

$= \frac{2}{\cos A}$

We know that $\frac{1}{\cos A} = \sec A$.

$= 2 \sec A$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 3 — Proving identity (iii)

Let's start with the left-hand side (LHS) of the identity. We will write all terms in terms of $\sin \theta$ and $\cos \theta$.

$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

$= \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$

Simplify the denominators.

$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$

Invert and multiply the fractions.

$= \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \theta}{\sin \theta - \cos \theta} + \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\cos \theta - \sin \theta}$

$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$

Notice that $(\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta)$.

$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$

Now, find a common denominator, which is $\sin \theta \cos \theta (\sin \theta - \cos \theta)$.

$= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Difference of Cubes: $a^3 - b^3 = (a-b)(a^2+ab+b^2)$. Here $a = \sin\theta$, $b = \cos\theta$.

Use the difference of cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Here, $a = \sin \theta$ and $b = \cos \theta$.

$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Cancel out the common term $(\sin \theta - \cos \theta)$. Also, $\sin^2 \theta + \cos^2 \theta = \mathbf{1}$.

$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$

Separate the terms in the numerator.

$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$= \frac{1}{\sin \theta \cos \theta} + 1$

We know that $\frac{1}{\sin \theta} = \text{cosec } \theta$ and $\frac{1}{\cos \theta} = \sec \theta$.

$= \text{cosec } \theta \sec \theta + 1$

Rearrange the terms to match the RHS.

$= 1 + \sec \theta \text{ cosec } \theta$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 4 — Proving identity (iv)

Let's simplify the left-hand side (LHS) first. We will convert $\sec A$ into terms of $\cos A$.

$\frac{1 + \sec A}{\sec A}$

$= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$

Simplify the numerator.

$= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}}$

Invert and multiply.

$= \frac{\cos A + 1}{\cos A} \cdot \frac{\cos A}{1}$

$= \cos A + 1$

Now, let's simplify the right-hand side (RHS). Pythagorean Identity: $\sin^2 A = 1 - \cos^2 A = (1-\cos A)(1+\cos A)$.

We know that $\sin^2 A = 1 - \cos^2 A$.

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

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

Factor the numerator as a difference of squares: $1 - \cos^2 A = (1 - \cos A)(1 + \cos A)$.

$= \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A}$

Cancel out the common term $(1 - \cos A)$.

$= 1 + \cos A$

Since LHS = $1 + \cos A$ and RHS = $1 + \cos A$, they are equal.

$\boxed{\text{LHS} = \text{RHS}}$

Step 5 — Proving identity (v)

Let's start with the left-hand side (LHS) of the identity. We will divide the numerator and denominator by $\sin A$.

$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$

$= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$

Convert to $\cot A$ and $\text{cosec } A$.

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

Rearrange the terms in the numerator.

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

Pythagorean Identity: $\csc^2 A - \cot^2 A = 1$, which means $\csc^2 A - \cot^2 A = (\csc A - \cot A)(\csc A + \cot A)$ by difference of squares.

We are given the identity $\text{cosec}^2 A = 1 + \cot^2 A$, which means $\text{cosec}^2 A - \cot^2 A = \mathbf{1}$. Substitute 1 in the numerator with $\text{cosec}^2 A - \cot^2 A$.

$= \frac{(\cot A + \text{cosec } A) - (\text{cosec}^2 A - \cot^2 A)}{\cot A - \text{cosec } A + 1}$

Factor $\text{cosec}^2 A - \cot^2 A$ as $(\text{cosec } A - \cot A)(\text{cosec } A + \cot A)$.

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

Factor out $(\cot A + \text{cosec } A)$ from the numerator.

$= \frac{(\cot A + \text{cosec } A) [1 - (\text{cosec } A - \cot A)]}{\cot A - \text{cosec } A + 1}$

$= \frac{(\cot A + \text{cosec } A) [1 - \text{cosec } A + \cot A]}{\cot A - \text{cosec } A + 1}$

The term in the square brackets is identical to the denominator.

$= \cot A + \text{cosec } A$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 6 — Proving identity (vi)

Let's start with the left-hand side (LHS) of the identity. We will multiply the numerator and denominator inside the square root by $(1 + \sin A)$.

$\sqrt{\frac{1 + \sin A}{1 - \sin A}}$

$= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}}$

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

Pythagorean Identity: $1 - \sin^2 A = \cos^2 A$. Taking the positive square root is valid because A is acute.

We know that $1 - \sin^2 A = \cos^2 A$.

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

Take the square root of the numerator and denominator.

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

$= \frac{1 + \sin A}{\cos A}$

Separate the terms in the numerator.

$= \frac{1}{\cos A} + \frac{\sin A}{\cos A}$

We know that $\frac{1}{\cos A} = \sec A$ and $\frac{\sin A}{\cos A} = \tan A$.

$= \sec A + \tan A$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 7 — Proving identity (vii)

Let's start with the left-hand side (LHS) of the identity. We will factor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator.

$\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$

$= \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$

We know that $\sin^2 \theta = 1 - \cos^2 \theta$. Substitute this into the numerator.

$= \frac{\sin \theta (1 - 2(1 - \cos^2 \theta))}{\cos \theta (2 \cos^2 \theta - 1)}$

$= \frac{\sin \theta (1 - 2 + 2 \cos^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$

$= \frac{\sin \theta (2 \cos^2 \theta - 1)}{\cos \theta (2 \cos^2 \theta - 1)}$

Cancel out the common term $(2 \cos^2 \theta - 1)$.

$= \frac{\sin \theta}{\cos \theta}$

We know that $\frac{\sin \theta}{\cos \theta} = \tan \theta$.

$= \tan \theta$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 8 — Proving identity (viii)

Let's start with the left-hand side (LHS) of the identity. We will expand both squared terms using $(a+b)^2 = a^2 + 2ab + b^2$.

$(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$

$= (\sin^2 A + 2 \sin A \text{ cosec } A + \text{cosec}^2 A) + (\cos^2 A + 2 \cos A \sec A + \sec^2 A)$

Reciprocal products: $\sin A \cdot \csc A = 1$ and $\cos A \cdot \sec A = 1$ (each ratio times its reciprocal always equals 1).

We know that $\sin A \text{ cosec } A = \mathbf{1}$ and $\cos A \sec A = \mathbf{1}$.

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

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

Group $\sin^2 A + \cos^2 A$, which equals 1.

$= (\sin^2 A + \cos^2 A) + 2 + 2 + \text{cosec}^2 A + \sec^2 A$

$= 1 + 4 + \text{cosec}^2 A + \sec^2 A$

$= 5 + \text{cosec}^2 A + \sec^2 A$

Pythagorean Identities: $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$.

Now, use the identities $\text{cosec}^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$.

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

$= 5 + 1 + \cot^2 A + 1 + \tan^2 A$

$= 7 + \tan^2 A + \cot^2 A$

This is the right-hand side (RHS) of the identity.

$\boxed{\text{LHS} = \text{RHS}}$

Step 9 — Proving identity (ix)

Let's simplify the left-hand side (LHS) first. We will convert $\text{cosec } A$ and $\sec A$ into terms of $\sin A$ and $\cos A$.

$(\text{cosec } A - \sin A)(\sec A - \cos A)$

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

Simplify each parenthesis.

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

Pythagorean Identity: $1 - \sin^2 A = \cos^2 A$ and $1 - \cos^2 A = \sin^2 A$.

We know that $1 - \sin^2 A = \cos^2 A$ and $1 - \cos^2 A = \sin^2 A$.

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

Multiply the terms.

$= \frac{\cos^2 A \sin^2 A}{\sin A \cos A}$

Cancel out common terms.

$= \cos A \sin A$

Now, let's simplify the right-hand side (RHS). We will convert $\tan A$ and $\cot A$ into terms of $\sin A$ and $\cos A$.

$\frac{1}{\tan A + \cot A}$

$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

Find a common denominator in the denominator.

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

We know that $\sin^2 A + \cos^2 A = \mathbf{1}$.

$= \frac{1}{\frac{1}{\cos A \sin A}}$

Invert and multiply.

$= 1 \cdot (\cos A \sin A)$

$= \cos A \sin A$

Since LHS = $\cos A \sin A$ and RHS = $\cos A \sin A$, they are equal.

$\boxed{\text{LHS} = \text{RHS}}$

Step 10 — Proving identity (x)

This identity has three parts that must be equal. We will prove it in two steps.

Part 1: Proving $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$

Let's start with the left-hand side (LHS). Pythagorean Identities: $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \csc^2 A$.

We use the identities $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \text{cosec}^2 A$.

$\frac{1 + \tan^2 A}{1 + \cot^2 A}$

$= \frac{\sec^2 A}{\text{cosec}^2 A}$

Convert $\sec^2 A$ to $\frac{1}{\cos^2 A}$ and $\text{cosec}^2 A$ to $\frac{1}{\sin^2 A}$.

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

Invert and multiply.

$= \frac{1}{\cos^2 A} \cdot \frac{\sin^2 A}{1}$

$= \frac{\sin^2 A}{\cos^2 A}$

$= \tan^2 A$

This matches the rightmost part of the identity.

Part 2: Proving $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Let's start with the left-hand side (LHS). We will convert $\tan A$ and $\cot A$ into terms of $\sin A$ and $\cos A$.

$\left(\frac{1 - \tan A}{1 - \cot A}\right)^2$

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

Simplify the numerator and denominator inside the parenthesis.

$= \left(\frac{\frac{\cos A - \sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}}\right)^2$

Invert and multiply the fractions inside the parenthesis.

$= \left(\frac{\cos A - \sin A}{\cos A} \cdot \frac{\sin A}{\sin A - \cos A}\right)^2$

Notice that $(\sin A - \cos A) = -(\cos A - \sin A)$.

$= \left(\frac{\cos A - \sin A}{\cos A} \cdot \frac{\sin A}{-(\cos A - \sin A)}\right)^2$

Cancel out the common term $(\cos A - \sin A)$.

$= \left(\frac{\sin A}{-\cos A}\right)^2$

$= \left(-\frac{\sin A}{\cos A}\right)^2$

$= (-\tan A)^2$

$= \tan^2 A$

This also matches the rightmost part of the identity. Since both parts simplify to $\tan^2 A$, the identity is proven.

$\boxed{\text{LHS} = \text{RHS}}$

Answer

(i) $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ is proven. (ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$ is proven. (iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta$ is proven. (iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$ is proven. (v) $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$ is proven. (vi) $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$ is proven. (vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$ is proven. (viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ is proven. (ix) $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$ is proven. (x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$ is proven.