Math Notes For Class 11 Chapter 7 Ex 7.1 Solved & Quiz

Question 1 (Part i): Simplify the algebraic trigonometric expression: $\frac{\sin\theta + \tan\theta}{1 + \cos\theta}$.

Solution:
Step 1: Express $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$: $$\text{Numerator} = \sin\theta + \frac{\sin\theta}{\cos\theta}$$ Step 2: Factor out $\sin\theta$ from the numerator: $$\text{Numerator} = \sin\theta \left(1 + \frac{1}{\cos\theta}\right) = \sin\theta \left(\frac{\cos\theta + 1}{\cos\theta}\right)$$ Step 3: Substitute the rewritten numerator into the original rational expression: $$\frac{\sin\theta \left(\frac{1 + \cos\theta}{\cos\theta}\right)}{1 + \cos\theta}$$ Step 4: Cancel the common factor $(1 + \cos\theta)$ present in both numerator and denominator (assuming $\cos\theta \neq -1$): $$= \frac{\sin\theta}{\cos\theta} = \tan\theta$$
Answer: $$\tan\theta$$

Question 1 (Part ii): Simplify the expression: $\frac{\cos\theta}{1 - \sin\theta} - \frac{\cos\theta}{1 + \sin\theta}$.

Solution:
Step 1: Take the Least Common Multiple (LCM) of the denominators, which is $(1 - \sin\theta)(1 + \sin\theta)$: $$\text{LCM} = 1 - \sin^2\theta = \cos^2\theta$$ Step 2: Combine the fractions over the common denominator: $$E = \frac{\cos\theta(1 + \sin\theta) - \cos\theta(1 - \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}$$ Step 3: Expand the numerator: $$\text{Numerator} = \cos\theta + \cos\theta\sin\theta - \cos\theta + \cos\theta\sin\theta = 2\cos\theta\sin\theta$$ Step 4: Divide by the simplified denominator $\cos^2\theta$: $$E = \frac{2\cos\theta\sin\theta}{\cos^2\theta} = 2 \cdot \frac{\sin\theta}{\cos\theta} = 2\tan\theta$$
Answer: $$2\tan\theta$$

Question 2: Prove the trigonometric identity: $\frac{1 - \sin\theta}{\cos\theta} = \frac{\cos\theta}{1 + \sin\theta}$.

Solution:
Step 1: Start with the Left-Hand Side (LHS): $$\text{LHS} = \frac{1 - \sin\theta}{\cos\theta}$$ Step 2: Multiply the numerator and denominator by the conjugate of the numerator, $(1 + \sin\theta)$: $$\text{LHS} = \frac{(1 - \sin\theta)(1 + \sin\theta)}{\cos\theta(1 + \sin\theta)}$$ Step 3: Apply the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ in the numerator: $$\text{Numerator} = 1 - \sin^2\theta = \cos^2\theta$$ Step 4: Substitute back into the expression and simplify: $$\text{LHS} = \frac{\cos^2\theta}{\cos\theta(1 + \sin\theta)} = \frac{\cos\theta}{1 + \sin\theta} = \text{RHS}$$ Since LHS = RHS, the identity is verified.
Answer: $$\frac{\cos\theta}{1 + \sin\theta}$$

Question 3: Prove that $(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta$.

Solution:
Step 1: Consider the Left-Hand Side (LHS) and convert to sine and cosine: $$\text{LHS} = \left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right)^2$$ Step 2: Combine the terms inside the parentheses using LCM: $$\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$$ Step 3: Square the result: $$\text{LHS} = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 = \frac{1}{\sin^2\theta\cos^2\theta}$$ Step 4: Consider the Right-Hand Side (RHS) and express in sine and cosine: $$\text{RHS} = \sec^2\theta + \csc^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta}$$ Step 5: Take the LCM of the RHS: $$\text{RHS} = \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta\sin^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$$ Since $\text{LHS} = \text{RHS} = \frac{1}{\sin^2\theta\cos^2\theta}$, the identity is proven.
Answer: $$\sec^2\theta + \csc^2\theta$$

Question 4 (Part i): Find the exact value of $\sin(105^\circ)$ without using a calculator.

Solution:
Step 1: Decompose $105^\circ$ into standard known angles: $105^\circ = 60^\circ + 45^\circ$.
Step 2: Apply the sine sum formula $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$: $$\sin(105^\circ) = \sin(60^\circ + 45^\circ) = \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ)$$ Step 3: Substitute exact values ($\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 60^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$): $$\sin(105^\circ) = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)$$ Step 4: Multiply and combine fractions: $$\sin(105^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$
Answer: $$\frac{\sqrt{6} + \sqrt{2}}{4}$$

Question 4 (Part ii): Find the exact value of $\cos(75^\circ)$ without using a calculator.

Solution:
Step 1: Express $75^\circ$ as $45^\circ + 30^\circ$.
Step 2: Apply the cosine sum formula $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$: $$\cos(75^\circ) = \cos(45^\circ + 30^\circ) = \cos(45^\circ)\cos(30^\circ) - \sin(45^\circ)\sin(30^\circ)$$ Step 3: Substitute exact values ($\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 30^\circ = \frac{1}{2}$): $$\cos(75^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$$ Step 4: Combine terms: $$\cos(75^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$
Answer: $$\frac{\sqrt{6} - \sqrt{2}}{4}$$

Question 5: Prove that $\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ$.

Solution:
Step 1: Consider the Left-Hand Side (LHS) and divide both numerator and denominator by $\cos 11^\circ$: $$\text{LHS} = \frac{\frac{\cos 11^\circ}{\cos 11^\circ} + \frac{\sin 11^\circ}{\cos 11^\circ}}{\frac{\cos 11^\circ}{\cos 11^\circ} - \frac{\sin 11^\circ}{\cos 11^\circ}} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ}$$ Step 2: Recall that $\tan 45^\circ = 1$. Rewrite the constant $1$ as $\tan 45^\circ$: $$\text{LHS} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - (\tan 45^\circ)(\tan 11^\circ)}$$ Step 3: Recognize this as the expansion of $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$ with $\alpha = 45^\circ$ and $\beta = 11^\circ$: $$\text{LHS} = \tan(45^\circ + 11^\circ) = \tan 56^\circ = \text{RHS}$$
Answer: $$\tan 56^\circ$$

Question 6: Prove the formula identity $\frac{\sin(\alpha+\beta)}{\cos\alpha \cos\beta} = \tan\alpha + \tan\beta$.

Solution:
Step 1: Start with the Left-Hand Side (LHS) and expand $\sin(\alpha+\beta)$: $$\text{LHS} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta}$$ Step 2: Split the single fraction into two separate fractions over the common denominator: $$\text{LHS} = \frac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta} + \frac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}$$ Step 3: Cancel $\cos\beta$ in the first term and $\cos\alpha$ in the second term: $$\text{LHS} = \frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}$$ Step 4: Use the quotient identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$: $$\text{LHS} = \tan\alpha + \tan\beta = \text{RHS}$$
Answer: $$\tan\alpha + \tan\beta$$